Hc Verma I for Class 12 Science Physics Chapter 10 - Rotational Mechanics

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Page No 192:

Answer:

Yes, such motion is possible if the translation takes place along the axis of rotation. This type of motion is called screw motion.

Page No 192:

Answer:

Yes, it is an example of pure rotation.
The axis of rotation passes through the pivot of the pendulum, perpendicular to the plane containing the pendulum.

Page No 192:

Answer:

No, we cannot use componendo-dividendo theorem of proportion here. This is because $α$ and a, and v and $ω$ are dimensionally different. Therefore, v + $ω$ and/or $α$ + a are not possible.

Page No 192:

Answer:

Yes, there is an angular rotation of the ball about its centre.
Yes, angular velocity of the ball about its centre is same as the angular velocity of the ball about the fixed point.

Explanation:
Let the time period of angular rotation of the ball be T.
Therefore, we get:
Angular velocity of the ball about the fixed point = $\frac{2 π}{T}$
After one revolution about the fixed centre is completed, the ball has come back to its original position. In this case, the point at which the ball meets with the string is again visible after one revolution. This means that it has undertaken one complete rotation about its centre.
The ball has taken one complete rotation about its centre. Therefore, we have:

Angular displacement of the ball = $2 π$
Time period = T
So, angular velocity is again $\frac{2 π}{T}$ . Thus, in both the cases, angular velocities are the same.

Page No 192:

Answer:

No, we cannot see the other face of the Moon from the Earth.

Yes, a person on the Moon can see all the faces of the Earth.

Explanation: Angular velocity of the Moon about its own axis of rotation is same as its angular velocity of revolution about the Earth. This means that its rotation time period equals its revolution time period. So, we can see only one face of the Moon from the Earth.

However, angular velocity of the Earth about its axis is not same as the angular velocity of Moon about the Earth. So, all the faces of the Earth is visible from the Moon.

Page No 192:

Answer:

No, its not always correct.

     Figure 1
Explanation: If the centre of mass of the body is not on the same vertical line as the normal reaction R of the body, a net torque acts on the body about its vertical axis. In fig. 1, R and CM lies in the same vertical line. Thus, there is no torque about any vertical axis

      Figure 2

But in fig. 2, as R and CM do not lie along the same vertical line, there exists a torque about the vertical axis.

Page No 192:

Answer:

$No, \underset{r}{\to} \times \underset{τ}{\to} | | \underset{Γ}{\to} is not true . In fact, it is never true . This is because : \underset{r}{\to} \times \underset{τ}{\to} = \underset{r}{\to} \times (\underset{r}{\to} \times \underset{F}{\to}) Applying vector triple product, we get : \underset{r}{\to} \times (\underset{r}{\to} \times \underset{F}{\to}) = (\underset{r}{\to} . \underset{F}{\to}) \underset{r}{\to} - (\underset{r}{\to} . \underset{r}{\to}) \underset{F}{\to} ∵ \underset{r}{\to} . \underset{r}{\to} = r^{2} = (\underset{r}{\to} . \underset{F}{\to}) \underset{r}{\to} - r^{2} \underset{F}{\to} If \underset{r}{\to} . \underset{F}{\to} = 0; that is, \underset{r}{\to} ⊥ \underset{F}{\to}, then : \underset{r}{\to} \times \underset{Γ}{\to} = - r^{2} \underset{F}{\to} We know that r^{2} is never negative and \underset{r}{\to} \times \underset{Γ}{\to} = - r^{2} \underset{F}{\to .} This implies that both vectors may be antiparallel to each other but not parallel .$

Page No 192:

Answer:

$We know that : \vec{τ} = \vec{r} \times \vec{F} Given : \vec{r} = - 0.5 \hat{i} m \vec{F} = - mg \hat{j} The torque becomes : \vec{τ} = 0.5 (- \hat{i}) \times m g (- \hat{j}) \vec{τ} = 0.5 mg \hat{k} [∵ \hat{i} \times \hat{j} = \hat{k}] No, there will be no angular acceleration on the particle due to the torque . Angular acceleration is given by α = \frac{τ}{I} . As the particle here moves in a straight line, the centre of rotation lies at a distance infinity (r = \infty); so, moment of inertia (I = m r^{2}) of the particle is infinity . ∴ α = 0$

Page No 192:

Answer:

$Let \vec{f_{1}}, \vec{f_{2}}, \vec{f_{3}}, . . . \vec{f_{n}} be the forces acting on a point P . Let O be the point along which torques (moments) will be taken . Let : \vec{f_{1}} + \vec{f_{2}} + \vec{f_{3}} + . . . + \vec{f_{n}} = \vec{R} . . . (1) Moments of force (torque) \vec{f_{i}} about O will be : \vec{τ_{1}} = \vec{OP} \times \vec{f_{1}} The sum of the torques about O will be : \vec{M} = \vec{OP} \times \vec{f_{1}} + \vec{OP} \times \vec{f_{2}} + . . . + \vec{OP} \times \vec{f_{n}} \Rightarrow \vec{M} = \vec{OP} \times (\vec{f_{1}} + \vec{f_{2}} + \vec{f_{3}} + . . . + \vec{f_{n}}) \Rightarrow \vec{M} = \vec{OP} \times \vec{R} [From (1)] Thus, we see that the torque of the resultant force \vec{R} of the forces \vec{f_{1}}, \vec{f_{2}}, \vec{f_{3}}, . . . \vec{, f_{n}} gives the sum of the moments of the torques .$

Page No 192:

Answer:

$No, if the sum of all the forces acting on a body is z ero, the body is not necessarily in equilibrium . To be in equilibrium, the sum of torque acting on the body must be z ero too (s ee the fig .) . In the above case, although the sum of the forces acting on the body is zero (\vec{F_{1}} + (- \vec{F_{1}}) = 0) . Still, the body will rotate along \vec{OP} . So, it won' t remain in equilibrium .$

Page No 192:

Answer:

No, angular momentum is dependent on the position vector of the particle, angle between the radius vector and the linear velocity of the particle. So, there may be finite angular momentum along any different point even if it is zero at a particular point.If angular momentum is zero along O' but finite along O.

Page No 192:

Answer:

$No, it is not necessary that the torque about any other point be zero if it is zero about one point .$
$Let \vec{F} be the resultant force due to all the forces acting on the plane of the body . Therefore, torque due to force \vec{F} at any point will be the resultant torque . Now, we see that the torque due to \vec{F} at point Q will be zero because Q lies on the line of support of the force F but the torque due to force \vec{F} will not be zero along the point P .$

Page No 192:

Answer:

$Yes, if the torque due to forces in translation equillibrium is zero about a point, it will be zero about other point in the plane .$

$Let us consider a planner lamina of some mass, acted upon by forces \vec{F_{1}}, \vec{F_{2}}, . . . \vec{F_{i}}, etc . Let a force \vec{F_{i}} act on a i^{th} particle and torque due to \vec{F_{i}} be zero at a point Q . Since the body is in translation equillibrium, we have : Σ \vec{F_{i}} = 0 . . . (1) Again, torque about P is zero . Therefore, we have : Σ (\vec{r_{p i}} \times \vec{F_{i}}) = 0 . . . (2) Now, torque about point Q will be : Σ {\vec{r}}_{Qi} \times \vec{F_{i}} = Σ ({\vec{r}}_{Q P} + {\vec{r}}_{P i}) \times \vec{F_{i}} [From fig . 13] = Σ ({\vec{r}}_{Q P} \times \vec{F_{i}} + \vec{r_{P i}} \times \vec{F_{i}}) = Σ \vec{r_{Q P}} \times \vec{F_{i}} + Σ \vec{r_{P i}} \times \vec{F_{i}} = \vec{r_{QP}} \times Σ \vec{F_{i}} + 0 [From (2)] = \vec{r_{Q P}} \times 0 [From (1)] = 0 Thus, \vec{F} is zero about any other point Q .$

Page No 192:

Answer:

$The centre of mass (CM) of a rectangular block lies in the middle of the block . When the block is projected less than half of its length (CM being over the table), no net force acts on it . Thus, no net torque acts upon the body . But if the block is projected more than half of its length outside the table (CM being outside the table), gravitational force acts along the CM of the block . This force produces a moment along the edge of the table . This rotates the block, and as a result, it falls down .$

Page No 192:

Answer:

$When the man tries to touch his toe, he exerts force along the hand downwards . This force produces a moment along the Centre of Mass (CM) of the man as shown in the figure . This moment makes him rotate and, thus, he falls down after losing the balance .$

Page No 192:

Answer:

$The ladder is more likely to slide when the man stands near the top . This is because when the man stands near the top, it creates more torque compared to the torque caused by the weight of man near the bottom .$

$When the man stands near the bottom, the Centre of Gravity of the ladder is shifted to C' from C . Now, the couple due to forces (m + M) g and N makes the ladder fall . We see that due to its shift from C to C', the moment arm of the couple decreases from r to r'; hence, the couple decreases .$

$When the man stands near the top of the ladder, the Centre of Mass shifts from C to C' . This increases the moment arm of the couple and from r to r' . Increase in moment arm increases the couple and, thus, the ladder easily falls .$

Page No 192:

Answer:

When the balance is kept at an angle, there is a net extra torque given to one of its arm. When the extra torque is removed, the balance becomes torque free and sum of all the torque acting on it is zero.

But balance kept at an angle has got a greater potential energy compared to the balance kept horizontal. The potential energy acquired is due to the initial torque applied on it. This displaces the balance by an angle. As soon as the body is set free to rotate, the body tends to have the lowest potential energy. Thus, potential energy starts converting in to kinetic energy, but on the other side, kinetic energy converts into potential energy when the other arm of the balance is raised. This energy transformation oscillates the balance. But in this process, friction with the air and fulcrum dissipates energy converting into heat. Finally, the balance loses the energy and becomes horizontal, or attains equilibrium.

Page No 192:

Answer:

It will require more force to set the bar into rotation by clamping at A and then clamping at B.

Explanation: Since the rod has mass density increasing towards B, the Center of Mass (CM) of the rod is near B. If the rod is clamped along A, the distance of CM of the rod from the pivot will be greater when the rod is clamped along B. Greater distance of CM from the Center of rotation increases the moment of inertia of the rod and hence more torque will be necessary to rotate the bar about A. Greater torque implies greater force will be necessary to rotate it.
F_A = Force required to rotated the rod clamped at A
R_A= Distance of CM from pivot A
M = Mass of the rod
F_B= Force required to rotate the rod clamped at B
R_B= Distance of CM from pivot B
We have R_A>R_B.
We have to find the torque required to rotate rod clamped at A to produce angular acceleration a.
T_A = MR_A²a = R_AF_A
=> F_A = MR_Aa
We have to find torque required to rotate rod clamped at B to produce angular acceleration a.
T_B = MR_B²a = R_BF_B
=> F_B = MR_Ba
On comparing, since R_A>R_B,we get:
F_A>F_B

Page No 192:

Answer:

Yes, because tall buildings have their CM much above the ground. It increases moment of inertia of the Earth. As the Earth’s rotation does not involve torque, its angular momentum is constant. Thus, an increase in MI leads to lower angular velocity of the Earth about its axis of rotation. This means length of night and day will increase. However, the increase is very small.

Page No 192:

Answer:

Ice caps near the poles concentrate the mass of water at the poles through which axis of rotation of the Earth passes. If the ice melts, water will spread across the globe due to hydrostatic equilibrium and tend to move to the equatorial areas of the Earth due to centrifugal force of rotation. Mass, now being distributed more along the equator, will increase MI of the Earth and this in turn will decrease the angular velocity of the Earth. Decrease in angular velocity will increase the duration of day-night.

Page No 192:

Answer:

The body with the smallest moment of inertia will roll down taking the smallest time. Here, the solid sphere has the lowest moment of inertia among all the other bodies. So, it will roll down taking the least time.

Page No 192:

Answer:

Some points on the equator of the sphere has got vertical velocity with respect to the direction of motion of the sphere.

Page No 193:

Answer:

(c) 0

For a purely rotating body, the axis of rotation is always perpendicular to the velocity of the particle.
Therefore, we have:
$\vec{A} . \vec{B} = 0$

Page No 193:

Answer:

(c) 0

The unit vector along the axis of rotation and the unit vector along the resultant force on the particle are perpendicular to each other in a uniform rotation.
Therefore, we have:
$\vec{A} . \vec{B} = 0$

Page No 193:

Answer:

(b) remains constant

For angular momentum, we have:
$\vec{L} = m (\vec{r} \times \vec{v}) \vec{v} = v \hat{i} and \vec{r} = x \hat{i} + y \hat{j} So, \vec{L} = - m v y \hat{k}$
m, v and y are constant; therefore, angular momentum remains constant.

Page No 193:

Answer:

(d) $ω$ is independent of r

In a pure rotation, angular velocity of all the particles remains same and does not depend on the position of the particle from the axis of rotation.

Page No 193:

Answer:

(c) y = 2 x

It is given that angular velocity is same for both the wheels.
Therefore, we have:
v_A= $ω$ R
v_B = $ω$ 2R
x = v_At = $ω$ Rt ... (i)
y = v_Bt = $ω$ (2R)t ... (ii)
From equations (i) and (ii), we get:
y = 2 x

Page No 193:

Answer:

(c) horizontal and intersecting the axis

Because resultant force on a particle of the body rotating uniformly is always perpendicular to the rotation axis and pass through it.

Page No 193:

Answer:

(b) horizontal and skew with the axis

The resultant force on a particle of the body rotating non-uniformly is always horizontal and skew with the rotation axis because net torque on the body is non-zero.

Page No 193:

Answer:

(a) $\vec{r} . \vec{Γ} = 0 and \vec{F} . \vec{Γ} = 0$

We have:
$\vec{τ} = \vec{r} \times \vec{F}$
Thus, $\vec{τ}$ is perpendicular to $\vec{r}$ and $\vec{F}$ .
Therefore, we have:
$\vec{r} . \vec{τ} = 0 and \vec{F} . \vec{τ} = 0$

Page No 193:

Answer:

(d) $\frac{1}{2} m ω^{2} l$ .

Consider a small portion of rod at a distance x from the clamped end (as shown in fig.) with width dx and mass dm.

Centripetal force on this portion = $ω^{2} x d m$
$d m = (\frac{m}{l}) l d x$
Force on the whole rod = F = $\int_{0}^{l} ω^{2} x \frac{m}{l} d x$

$∴$ F = $\frac{1}{2} m ω^{2} l$

Page No 193:

Answer:

(c) at a distance l/2 from O.

It is given that there is no force along x-axis.
COM of rod will remain and will not shift along x-axis (horizontal direction).
Force gravity is acting along y-axis (vertical direction). So, COM will shift along the y-axis by l/2 distance and COM of horizontal rod is at a distance l/2 from one end. Therefore, lower end of the rod will remain at a distance l/2 from O.

Page No 193:

Answer:

(c) I_A < I_B

Moment of inertia of circular disc of radius r:
I = $\frac{1}{2} m r^{2}$
Mass = Volume × Density
Volume of disc = $π r^{2} t$
Here, t is the thickness of the disc.
As density is same for both the rods, we have:
Moment of inertia, $I \propto thickness \times {(radius)}^{4}$

$\frac{I_{A}}{I_{B}} = \frac{t . {(r)}^{4}}{\frac{t}{4} {(4 r)}^{4}} < 1 \Rightarrow \frac{I_{A}}{I_{B}} < 1$
$\Rightarrow I_{A} < I_{B}$

Page No 193:

Answer:

(a) $ν_{A} > ν_{B}$

$τ = I α (magnitude)$
For equal torque, we have:
$I_{A} α_{A} = I_{B} α_{B}$

I_A < I_B
⇒ $α_{A} > α_{B}$ ...(i)

Now, $ω = α t$
Or, $\frac{v}{r} = α t$
$ν_{A} > ν_{B}$ (using (i))

Page No 193:

Answer:

(a) increases

Moment of inertia of a mass is directly proportional to the square of the distance of mass from the axis of rotation.
Therefore, we have:
$I \propto r^{2}$

As the tube is rotated, water is collected at the end of tube because of centrifugal force and distance from the rotation axis increases. Hence, moment of inertia increases.

Page No 193:

Answer:

(a) $M r^{2}$

Consider an element of length, dl = rdθ.
$d m = \frac{M}{π r} d l = \frac{M}{π r} r d θ$
MOI of semicircular wire = $\int_{0}^{π} r^{2} d m$
$I = \int_{0}^{π} r^{2} \frac{m}{π r} r d θ \Rightarrow I = m r^{2}$

Page No 193:

Answer:

(a) I₁ < I₂

In the given case, we have:
MOI $\propto$ Density
The density of iron is more; therefore, I₂ will be greater.

Page No 193:

Answer:

(d) cannot be deduced with this information.

I_x = m₂a² + m₃a² = 0.20    ...(i)
I_y = m₁a² + m₃a² = 0.20    ...(ii)
I_z= m₁a²+ m₂a²    ...(iii)
We have three equations and four variables. So, I_zcannot be deduced with the given information.

Page No 193:

Answer:

(d) $\frac{1}{2} M g a sinθ$

Let N be the normal reaction on the block.

From the free body diagram of the block, it is clear that the forces N and mgcosθ pass through the same line. Therefore, there will be no torque due to N and mgcosθ. The only torque will be produced by mgsinθ.
$∴ \vec{τ} = \vec{F} \times \vec{r} Since a is the edge of the cube, r = \frac{a}{2} . Thus, we have : τ = m g \sin θ \times \frac{a}{2} = \frac{1}{2} m g a \sin θ$

Page No 194:

Answer:

(b) $\frac{ω M}{M + 2 m}$

No external torque is applied on the ring; therefore, the angular momentum will be conserved.
$I ω = I' ω' \Rightarrow ω' = \frac{I ω}{I'} . . . (i)$
$I = M r^{2} I' = M r^{2} + 2 m r^{2}$
On putting these values in equation (i), we get:
$ω' = \frac{ω M}{M + 2 m}$

Page No 194:

Answer:

(c) remains unchanged

Rate of change of angular momentum of the body is directly proportional to the net external torque acting on the body.
No external torque is applied on the person or on the table; therefore, the angular momentum will be conserved.

Page No 194:

Answer:

(c) $\sqrt{2} ν_{0}$

For pure rolling, $ω r = v_{0}$

As shown in the figure, the velocity of the particle will be the resultant of v₀ and ωr.
Therefore, we have:
$v_{n e t} = \sqrt{{v_{0}}^{2} + {(ω r)}^{2}} v_{n e t} = \sqrt{{v_{0}}^{2} + {v_{0}}^{2}} v_{n e t} = \sqrt{2} v_{0}$

Page No 194:

Answer:

(a) along the velocity of the wheel

As the distance covered in one revolution about the centre is less than the perimeter of the wheel, it means that the direction of torque due to frictional force opposes the motion of wheel, i.e., the frictional force acting on the wheel by the surface is along the velocity of the wheel.
FIG.

Page No 194:

Answer:

(d) 0%.

On a frictionless road, we have:
Angular velocity of the engine = 0
Therefore, increase in petrol input will not affect the angular velocity and hence the linear velocity of the scooter will remain the same.

Page No 194:

Answer:

(d) all will take same time.

The incline is smooth; therefore, all bodies will slip on the incline. Also, as the mass of bodies is same, they will reach the bottom in equal time.

Page No 194:

Answer:

(d) all will take same time

Let θ be the inclination angle.

From the free body diagram, we have:
$N = m g \cos θ . . . (i) m a = m g \sin θ - f_{r} . . . (i i) Putting f_{r} = μ N in (i i) w e g e t, a = g (\sin θ - μ \cos θ)$

The friction coefficients between the objects and the incline are same and not sufficient to allow pure rolling; therefore, all the bodies come down with the same acceleration.

Page No 194:

Answer:

(b) the hollow sphere

Torque is same for all the bodies; therefore, the angular momentum will be conserved.
Now, total kinetic energy = $\frac{1}{2} m v^{2} + \frac{L^{2}}{2 I}$
So, the body with greater value of moment of inertia will have smallest kinetic energy at the bottom of the incline.

Order of the moment of inertia of the bodies:
hollow sphere > disc > solid sphere

Hence, the hollow sphere will have the smallest kinetic energy at the bottom.

Page No 194:

Answer:

(b) 2l

For pure rolling, $ω r = v_{0}$

Figure:

As shown in the figure, the velocity of the string will be resultant of v₀ and ωr.
$v_{n e t} = v_{0} + (ω r) v_{n e t} = 2 v_{0}$
Let:
Linear distance travelled by the cylinder in time (t), $v_{0} t = l$
∴ Linear distance travelled by the string in same time, $2 v_{0} t = 2 l$

Page No 194:

Answer:

(b) may pass through the centre of mass
(d) may pass through a particle of the body

It is not necessary that the axis of rotation of a purely rotating body should pass through the centre of mass or through a particle of the body. It can also lie outside the body.

Page No 194:

Answer:

(b) A is true but B is false.

In non-inertial frames, $\frac{d L}{d t} = Γ_{total}$
Here, $Γ_{total}$ is is the total torque on the system due to all the external forces acting on the system. So, equation (B) is not true as in non-inertial frames, pseudo force must be applied to study the motion of the object.

Page No 194:

Answer:

(b) is zero about a point on the straight line
(c) is not zero about a point away from the straight line
(d) about any given point remains constant

(b) Angular momentum = $m (\vec{r} \times \vec{v})$
If the point is on the straight line, $\vec{r} and \vec{v}$ will have the same direction and their cross product will be zero. Hence, angular momentum is zero.

(c) If the point is not on the straight line, $\vec{r} and \vec{v}$ will not have the same direction and their cross product will not be zero. Hence, angular momentum is non-zero.

(d) No external torque is applied on the body; therefore, its angular momentum about any given point remains constant.

Page No 194:

Answer:

(a) angular momentum
(b) linear momentum

${\vec{F}}_{e x t} = 0 \Rightarrow {\vec{τ}}_{e x t} = 0$
That is, the change in linear momentum and angular momentum is zero. This is because:
$\frac{d \vec{P}}{d t} = {\vec{F}}_{ext} And \frac{d \vec{L}}{d t} = {\vec{τ}}_{ext}$

Page No 194:

Answer:

(c) If the axes are parallel, I_A < I_B

If axes A and B are parallel, we get:
_{$I_{B} = I_{A} + m r^{2}$}
Here, r is the distance between two axes and m is the mass of the body.
$∴$ I_A < I_B

Page No 194:

Answer:

(b) The particles on the diameter mentioned above do not have any linear acceleration.

Explanation:
Linear acceleration of a rotating particle is given as:
$\vec{a} = \vec{r} \times \vec{α}$

(b)
The sphere is rotating about a diameter; therefore, the position vector of the particles on the diameter is zero. Thus, linear acceleration of the particle is zero.

(c) and (d)
All the particles of the body have the same angular velocity. All the particle on the surface have different linear speeds that depend on the position of the particle from the axis of rotation.

Page No 195:

Answer:

Given:
t = 4 s
Initial angular velocity = $ω_{0} = 0$
Final angular velocity = $ω = 100 rev / s$
$ω = ω_{0} + α t α = \frac{ω}{t} α = \frac{100}{4} rev / s^{2} = 25 rev / s^{2} Now, we have : θ = ω_{0} t + \frac{1}{2} α t^{2} \Rightarrow θ = \frac{1}{2} \times 25 \times 16 = 200 ° \Rightarrow θ = 200 \times 2 π radians = 400 π radians$

Page No 195:

Answer:

Given:
Angular displacement of the wheel = $θ = 50 \times 2 π = 100 π$
Initial angular velocity of the wheel = $ω_{0} = 0$
After, t = 5 seconds
$θ = ω_{0} t + \frac{1}{2} α t^{2} \Rightarrow 100 π = \frac{1}{2} \times α \times (5)^{2} \Rightarrow 100 π = \frac{1}{2} \times α \times 25 \Rightarrow α = 8 π rad / s^{2} or 4 rev / s ω = ω_{0} + 2 α t \Rightarrow ω = 0 + 8 π \times 5 = 40 π rad / s \Rightarrow ω = 20 rev / s$

Page No 195:

Answer:

It is given that the area under the $ω - t$ curve gives the total angular displacement.
∴ Maximum angular velocity = $ω = α t$
$ω = 4 \times 10 = 40 rad / s$

Area under the curve
$= \frac{1}{2} \times 10 \times 40 + 40 \times 10 + \frac{1}{2} \times 40 \times 10 = 800 rad$
∴ Total angle rotated in 30 s = 800 rad.

Page No 195:

Answer:

(d) torque of the applied force

The torque of the applied force does not depend on the density of a rod. It depend on the distance between the pivot and the point where F is applied. So, it does not depend on which end of the rod is pivoted.

Page No 195:

Answer:

(a) the speed of the particle A is zero
(c) the speed of C is 2 $v_{0}$
(d) the speed of B is greater than the speed of O

For pure rolling, $ω r = v_{0}$
Velocity of the particle at A, B, C and D will be resultant of v₀ and ωr.

At point B,
$v_{n e t} = \sqrt{{v_{0}}^{2} + {(ω r)}^{2}} v_{n e t} = \sqrt{{v_{0}}^{2} + {v_{0}}^{2}} v_{n e t} = \sqrt{2} v_{0}$

At point C,
$v_{n e t} = v_{0} + (ω r) v_{n e t} = 2 v_{0}$

At point A,
$v_{n e t} = v_{0} - (ω r) v_{n e t} = 0$

At point O,
r = 0
$∴ v_{n e t} = v_{0}$

Page No 195:

Answer:

(c) the two spheres reach the bottom together

Acceleration of a sphere on the incline plane is given by:
$a = \frac{g \sin θ}{1 + \frac{I_{C O M}}{m r^{2}}} I_{C O M} for a solid sphere = \frac{2}{5} m r^{2} So, a = \frac{g \sin θ}{1 + \frac{2 m r^{2}}{5 m r^{2}}} = \frac{5}{7} g \sin θ$
a is independent of mass and radii; therefore, the two spheres reach the bottom together.

Page No 195:

Answer:

(b) The solid sphere reaches the bottom with greater speed.

Acceleration of a sphere on the incline plane is given by:
$a = \frac{g \sin θ}{1 + \frac{I_{C O M}}{m r^{2}}} I_{C O M} for a solid sphere = \frac{2}{5} m r^{2} So, a = \frac{g \sin θ}{1 + \frac{2 m r^{2}}{5 m r^{2}}} = \frac{5}{7} g \sin θ$
$I_{C O M} for a hollow sphere = \frac{2}{3} m r^{2} So, a' = \frac{g \sin θ}{1 + \frac{2 m r^{2}}{3 m r^{2}}} = \frac{3}{5} g \sin θ$
The acceleration of the solid sphere is greater; therefore, it will reach the bottom with greater speed.

Page No 195:

Answer:

(a) a smooth horizontal surface
(b) a smooth inclined surface

A sphere cannot roll on a smooth inclined surface and on a smooth horizontal surface because there is no backward force (force of friction) to prevent its slipping.

Page No 195:

Answer:

(a) on the rear wheels is in the forward direction
(b) on the front wheels is in the backward direction
(c) on the rear wheels has larger magnitude than the friction on the front wheels

Explanation:
(a) On the rear wheels, friction force is in the forward direction because it favours the motion and accelerates the car in forward direction.
(b) Because of the movement of the car in forward direction, front wheels push the road in forward direction and in reaction, the road applies friction force in the backward direction.
(c) As the car is moving in forward direction, the rear wheels have larger magnitude of friction force (in forward direction) than on the front wheels.

Page No 195:

Answer:

(c) it will translate and rotate about the centre

The given coefficient of friction ( $\frac{1}{7} g tanθ$ ) is less than the coefficient friction ( $\frac{2}{7} g tanθ$ ) required for perfect rolling of the sphere on the inclined plane. Therefore, sphere may slip while rolling and it will translate and rotate about the centre.

Page No 195:

Answer:

(a) decrease the linear velocity
(b) increase the angular velocity

If a sphere is rolled on a rough horizontal surface, the force of friction tries to oppose the linear motion and favours the angular motion.

Page No 195:

Answer:

(a) it will continue pure rolling

From the free body diagram of sphere, we have:
Net force on the sphere along the incline,
F_net = mgsinθ − macosθ ...(i)
On putting a = gtanθ in equation (i), we get:
F_net = 0
Therefore, if the sphere is set in pure rolling on the incline, it will continue pure rolling.

Page No 196:

Answer:

Given:
Angular acceleration of the body = $α = 2 rad / s^{2}$
Initial angular velocity of the body = $ω_{0} = 5 rad / s$
Final angular velocity of the body = $ω = 15 rad / s$
$We know that : ω = ω_{0} + α t \Rightarrow t = \frac{(ω - ω_{0})}{α} = \frac{(15 - 5)}{1} = 10 s Also, θ = ω_{0} t + \frac{1}{2} α t^{2} \Rightarrow θ = 5 \times 10 + \frac{1}{2} \times 1 \times 100 \Rightarrow θ = 50 + 50 = 100 rad$

Page No 196:

Answer:

Angular displacement of the body,
$θ = 5 rev = 10 π rad α = 2 rev / s^{2} = 4 π rad / s^{2} Initial angular velocity, ω_{0} = 0 Final angular velocity, ω = ? ω^{2} = (2 αθ) \Rightarrow ω = \sqrt{2 \times 4 π \times 10 π} = 4 π \sqrt{5} rad / s \Rightarrow ω = 2 \sqrt{5} rev / s$

Page No 196:

Answer:

Radius of disc = r = 10 cm = 0.1 m
Angular velocity of the disc = ω = 20 rad/s
∴ Linear velocity of point on the rim = $v = ω r$
$\Rightarrow v = 20 \times 0.1 = 2 m / s$

∴ Linear velocity of point at the middle of radius $v = \frac{ω r}{2} = 20 \times (\frac{0.1}{2}) = 1 m / s$

(i) (ii)

Page No 196:

Answer:

Angular acceleration of the disc, $α = 4 rad / s^{2}$
Distance of the particle from the axis of rotation, $r = 1 cm = 0.01 m$
$So, ω = α t = 4 rad / s (t = 1 s)$
Radial acceleration,
$α_{r} = ω^{2} r = 4^{2} \times 0.01 = 0.16 m / s^{2} = 16 cm / s^{2}$
Tangential acceleration,
$α_{T} = α r = 0.04 m / s^{2} = 4 cm / s^{2}$

Page No 196:

Answer:

It is given that the string is moving on the rim of the disc and block is connected with the string.
Therefore, the speeds of the block going down and the rim will be same.
Angular speed of the disc, ω = 10 rad/s
Radius of the pulley, r = 20 cm

Linear velocity of the rim, v = Tangential velocity = rω
v = 10 × 20 = 200 cm/s = 2 m/s
Therefore, velocity of the block = 2 m/s

Page No 196:

Answer:

(a) The distance of mass at A from the axis passing through side BC,
$(AD) = \frac{\sqrt{3}}{2} \times 10 = 5 \sqrt{3} cm$

Therefore, we have:
Moment of inertia of mass about the axis BC,
$l = m r^{2} = 200 \times {(5 \sqrt{3})}^{2} = 200 \times 25 \times 3 = 15000 gm - {cm}^{2} = 1.5 \times 10^{- 3} kg - m^{2}$

(b) Now, let the axis of rotation passes through A and is perpendicular to the plane of triangle.
Therefore, we have:
Net moment of inertia,
$l = m r^{2} + m r^{2} = 2 m r^{2} = 2 \times 200 \times 10^{2} = 400 \times 100 = 40000 gm - {cm}^{2} = 4 \times 10^{- 3} kg - m^{2}$

Page No 196:

Answer:

It is given that the perpendicular bisector of the metre scale is passed through the 50^th particle.
Therefore, on the L.H.S. of the axis, there will be 49 particles and on the R.H.S., there will be 50 particles.
Consider the two particles positioned at 49 cm and 51 cm.
Moment of inertia due to these two particles = 49 × (1)² + 51 × (1)²
I₁ = 100 × 1 = 100 gm-cm²
Similarly, if we consider particles positioned at 48 cm and 52 cm, we get:
I₂= 100 × (2)² gm-cm².
Thus, we will get 49 such sets and one particle at 100 cm. Therefore, total moment of inertia,
$I = (I_{1} + I_{2} + I_{3} . . . . . + I_{49}) + I'$
Here, I' is the moment of inertia of particle at 100 cm.
$So, I = 100 (1^{2} + 2^{2} + 3^{2} + . . . + 49^{2}) + 100 {(50)}^{2} = 100 (1^{2} + 2^{2} + 3^{2} + . . . + 50^{2}) = 100 \times \frac{(50 \times 51 \times 101)}{6} = 100 \times 25 \times 17 \times 101 = 4292599 gm - {cm}^{2}$
Or, I = 0.429 kg-m² ≃ 0.43 kg-m²

Page No 196:

Answer:

It is given that two bodies of mass m and radius r are moving along a common tangent.

Moment of inertia of the first body about XY tangent, $I' = I_{c o m} + m r^{2}$
$So, I' = \frac{2}{5} m r^{2} + m r^{2} = \frac{7}{5} m r^{2}$
Moment of inertia of the second body about XY tangent,
$I'' = \frac{2}{5} m r^{2} + m r^{2} = \frac{7}{5} m r^{2}$
Therefore, net moment of inertia,
$I = I' + I'' I = \frac{7}{5} m r^{2} + \frac{7}{5} m r^{2}$
$= \frac{14}{5} m r^{2} units$

Page No 196:

Answer:

Given:
Length of the rod, l = 1 m
Mass of the rod, m = 0.5 kg
Let the rod moves at a distance d from the centre.
On applying parallel axis theorem, we get:
Moment of inertia about that axis,
$I = (\frac{m l^{2}}{12}) + m d^{2} = 0.10$

$I = \frac{0.5 \times l^{2}}{12} + 0.5 \times d^{2} = 0.10 \Rightarrow \frac{1}{12} + d^{2} = 0.2 \Rightarrow d^{2} = 0.118 \Rightarrow d = 0.342 m from the centre$

Page No 196:

Answer:

Moment of inertia of the ring about a point on the rim of the ring and the axis perpendicular to the plane of the ring = mR² + mR² = 2mR² (from parallel axis theorem)
$We know that : m K^{2} = 2 m R^{2} (K = R adius of the gyration) \Rightarrow K = \sqrt{2 R^{2}} = \sqrt{2} R$

Page No 196:

Answer:

Moment of inertia of the disc about the centre and perpendicular to the plane of the disc = $\frac{1}{2}$ mr²
Radius of gyration of the disc about a point = Radius of the disc
$Therefore, m k^{2} = \frac{1}{2} m r^{2} + m d^{2}$
(k = Radius of gyration about acceleration point; d = Distance of that point from the centre)
$\Rightarrow K^{2} = \frac{r^{2}}{2} + d^{2} \Rightarrow r^{2} = \frac{r^{2}}{2} + d^{2} \Rightarrow \frac{r^{2}}{2} = d^{2} \Rightarrow d = \frac{r}{\sqrt{2}}$

Page No 196:

Answer:

Let there be a small sectional area of width dx at a distance x from the x-axis.

So,
Mass of element $= \frac{m}{a^{2}} \times a \times d x$
Moment of inertia about x-axis, $I_{x x} = 2 \int_{0}^{a / 2} \frac{m}{a^{2}} \times (a d x) \times x^{2}$
$\Rightarrow I_{x x} = 2 \times \frac{m}{a} {[\frac{x^{3}}{3}]}_{0}^{a / 2} = 2 \frac{m}{a} [\frac{a^{3}}{3 \times 8}] = \frac{m a^{2}}{12} Similarly, I_{y y} = \frac{m a^{2}}{12} Now, I_{z z} = I_{x x} + I_{y y} (Perpendicular axis theorem) \Rightarrow I_{z z} = 2 \times (\frac{m a^{2}}{12}) = \frac{m a^{2}}{6}$

The two diagonals are perpendicular to each other; therefore, we have:
$I_{z z} = I_{x' x'} + I_{y' y'} Also, I_{x x} = I_{y y} \Rightarrow I_{z z} = 2 I_{x' x'} \Rightarrow I_{x' x'} = \frac{m a^{2}}{12}$

Page No 196:

Answer:

Consider a ring of thickness dx at a distance r from the centre of the disc.
Mass of the ring , $d m = (A + B r) \times 2 π r \times d r$

Therefore, moment of inertia about the centre,
$I = \int_{0}^{a} r^{2} d m$
$= \int_{o}^{a} d (A + B r) 2 π r . d r \times r^{2} = \int_{o}^{a} 2 π A r^{3} d r + \int_{o}^{a} 2 π B r^{4} d r = {[2 π A (\frac{r^{4}}{4}) + 2 π B (\frac{r^{5}}{5})]}_{0}^{a} = 2 π (\frac{A a^{4}}{4} + \frac{B a^{5}}{5})$

Page No 196:

Answer:

Range of the particle $= (\frac{u^{2} \sin 2 θ}{g})$
At the highest point, we have:
Total force acting on the particle = mg (downward)
Distance between the line of force and the point of projection,
$\frac{(total range)}{2} = \frac{u^{2} \sin 2 θ}{2 g}$
$So, \vec{τ} = \vec{F} \times d_{⊥} = m g \times u^{2} \frac{\sin 2 θ}{2 g} \vec{τ} = m u^{2} \frac{\sin 2 θ}{2} = m u^{2} \sin θ \cos θ$

Therefore, the direction of torque is perpendicular to the plane of the motion.

Page No 196:

Answer:

Distance between the line of force and point of suspension, r = lsinθ
$Torque, \vec{τ} = \vec{F} \times \vec{r} \Rightarrow τ = w r sinθ = w l \sin θ$
Here, w is the weight of the bob.
The torque will be zero when the force acting on the body passes through the point of suspension, i.e., at the lowest point of suspension.

Page No 196:

Answer:

In the first case,
$τ_{1} = 6 \sin 30 ° \times (\frac{8}{100})$
In first case,
$τ_{2} = F \times (\frac{16}{100})$
To loosen the nut, torque in both the cases should be the same.
Thus, we have:
$τ_{1} = τ_{2}$
$\Rightarrow F \times \frac{16}{100} = 6 \sin 30 ° \times \frac{8}{100}$
$\Rightarrow F = \frac{(8 \times 3)}{16} = 1.5 N$

Page No 196:

Answer:

Torque about a point = Total force × Perpendicular distance
Let the anticlockwise torque and clockwise acting torque be positive and negative, respectively.
Torque at O due to 5 N force is zero as it is passing through O.
Torque at O due to 15 N force,
$τ_{1} = 15 \times 6 \times 10^{- 2} \times \sin 37 ° = 15 \times 6 \times 10^{- 2} \times \frac{3}{5} = 0.54 N - m (anticlockwise)$

Torque at O due to 10 N force,
$τ_{2} = 10 \times 4 \times 10^{- 2} = 0.4 N - m (clockwise)$

Torque at O due to 20 N force,
$τ_{3} = 20 \times 4 \times 10^{- 2} \times \sin 30 ° = 20 \times 4 \times 10^{- 2} \times \frac{1}{2} = 0.4 N - m (Anticlockwise)$

Resultant torque acting at O,
$τ = (0.54 - 0.4 + 0.4) = 0.54 N - m (anticlockwise)$

Page No 196:

Answer:

Let N be the normal reaction on the block.

From the free body diagram of the block, it is clear that forces N and mgcosθ pass through the same line. So, there will be no torque due to N and mgcosθ. The only torque will be produced by mgsinθ.
$∴ \vec{τ} = \vec{F} \times \vec{r} a is the edge of the cube . Therefore, we have : r = \frac{a}{2} ∴ τ = m g \sin θ \times \frac{a}{2} = \frac{1}{2} m g a \sin θ$

Page No 196:

Answer:

Torque about the centre due to force,
$\vec{τ} = \vec{F} \times \vec{r} = F \times \frac{L}{4} \times \sin 90 ° τ = F \times \frac{L}{4}$
Let the torque produces an angular acceleration α.
$τ = I α \Rightarrow τ = \frac{m L^{2}}{12} \times α (I of a rod = \frac{m L^{2}}{12}) \Rightarrow F \frac{L}{4} = \frac{m L^{2}}{12} \times α \Rightarrow α = \frac{3 F}{m L} Now, θ = \frac{1}{2} α t^{2} (initially at rest) \Rightarrow θ = \frac{3 F t^{2}}{2 m L}$

Page No 196:

Answer:

Consider a small area element of width dx and length a at a distance x from the axis of rotation.
Mass of element, dm $= \frac{m}{a^{2}} \times a d x$
(m = mass of the square, a = side of the plate)
$l = \int_{0}^{a} x^{2} d m l = \int_{0}^{a} \frac{m}{a^{2}} \times a x^{2} d x = \frac{m}{a} {[\frac{x^{3}}{3}]}_{0}^{a} = \frac{m a^{2}}{3}$

Now, torque produced,
$τ = (\frac{m a^{2}}{3}) \times α = \{\frac{120 \times 10^{- 3} \times 5^{2} \times 10^{- 4}}{3}\} \times . 02 = 40 \times 25 \times 10^{- 7} \times 0.2 = 2 \times 10^{- 5} N - m .$

Page No 196:

Answer:

Moment of inertia of a square plate about its diagonals,
I = $\frac{m a^{2}}{12}$
Therefore, we have:
Torque produced, $τ = I α$

$\Rightarrow τ = (\frac{m a^{2}}{12}) \times α = \frac{120 \times 10^{- 3} \times 5^{2} \times 10^{- 4}}{12} \times 0.2 = 10 \times 25 \times 10^{- 4} \times 0.2 = 0.5 \times 10^{- 5} N - m$

Page No 196:

Answer:

Let the angular deceleration produced due to frictional force be α.
Initial angular acceleration, $ω_{0} = 60 rad / s$
Final angular velocity, $ω = 0$
t = 5 min =300 s
We know that:
$ω = ω_{0} + α t$
$\Rightarrow α = - \frac{ω_{0}}{t} \Rightarrow α = - (\frac{60}{300}) = - \frac{1}{5} rad / s^{2}$

(a) Torque produced by the frictional force (R),
$τ = I α = 5 \times (\frac{- 1}{5}) = 1 N - m opposite to the rotation of wheel$

(b) By conservation of energy,
Total work done in stopping the wheel by frictional force = Change in energy
$W = \frac{1}{2} I ω^{2} = \frac{1}{2} \times 5 \times (60 \times 60) = 9000 joule = 9 kJ$

(c) Angular velocity after 4 minutes,

$ω = ω_{0} + α t = 60 - \frac{4 \times 60}{5} = \frac{60}{5} = 12 rad / s$
So, angular momentum about the centre,
$L = I ω = 5 \times 12 = 60 kg - m^{2} / s$

Page No 196:

Answer:

Rate of change of angular velocity, i.e., angular acceleration,
α = $(\frac{0.0016}{100}) rad / day$
$\Rightarrow α = \{\frac{0.0016}{{(86400)}^{2} \times 100 \times 365}\} [1 year = 365 days = 365 \times 86400 \sec]$
Torque produced by the ocean water in decreasing the Earth's angular velocity,

$τ = I α = \frac{2}{5} m r^{2} α = \frac{2}{5} \times 6 \times 10^{24} \times {(64 \times 10^{5})}^{2} \times \{\frac{0.0016}{86400^{2} \times 100 \times 365}\} = 5.8 \times 10^{20} N - m$

Page No 197:

Answer:

Initial angular velocity of the wheel,
$ω_{0} = 600 rpm = \frac{600}{60} = 10 revolutions per second$

After 10 seconds,
Final angular velocity of the wheel, $ω = 0$

$So, ω_{0} = - a t \Rightarrow α = - \frac{10}{10} = - 1 rev / s^{2} Now, t = 5 s We know that : ω' = ω_{0} + a t \Rightarrow ω' = 10 - 1 \times 5 = 5 rev / s$

Page No 197:

Answer:

Initial angular velocity of the wheel,
$ω = 100 rev / \min = \frac{100}{60} = \frac{5}{3} rev / s = \frac{10 π}{3} rad / s θ = 10 rev = 20 π rad r = 0.2 m$
Angular deceleration produced by the tangential force in order to stop the wheel after 10 revolutions, $α = \frac{ω^{2}}{2 θ}$
Torque by which the wheel will come to rest, $τ = I_{c m} α$
$Or, \Rightarrow τ = F r = I_{cm} α Putting the values of I_{cm} and α, we get : F r = \frac{1}{2} m r^{2} \times {(\frac{10 π}{3})}^{2} \times \frac{1}{2 \times 20 π} \Rightarrow F = \frac{1}{2} \times 10 \times 0.2 \times \frac{100 π^{2}}{(9 \times 2 \times 20 π)} \Rightarrow F = \frac{5 π}{18} = \frac{15.7}{18} = 0.87 N$

Page No 197:

Answer:

After t seconds, let the angular velocity of two cylinders be ω.
For the first cylinder,
$ω_{0} = 50 rev / s, α = 1 rev / s^{2}$
$∴ ω = 50 - α t \Rightarrow t = \frac{(ω - 50)}{- 1}$
For the second cylinder,
$ω_{0} = 0, α = 1 rev / s^{2}$
$∴ ω = α t \Rightarrow t = \frac{ω}{1}$
$On equating the value of t, we get : ω = (\frac{ω - 50}{- 1}) \Rightarrow 2 ω = 50 \Rightarrow ω = 25 rev / s ∴ t = \frac{25}{1} s = 25 s$

Page No 197:

Answer:

Given:
Initial angular acceleration, ω₀= 20 rad/s
Angular deceleration, $α = 2 rad / s^{2}$
$\Rightarrow t_{1} = \frac{ω_{0}}{α_{1}} = \frac{20}{2} = 10 s$
So, in 10 s the body will come to rest.
The same torque continues to act on the body, therefore, it will produce the same angular acceleration.
The kinetic energy of the body is same as the initial value; therefore, its angular velocity should be equal to the initial value ω₀.
Therefore, time required to come to that angular velocity,
$t_{2} = \frac{ω}{α} = \frac{20}{2} = 10 s$
So, total time required,
$t = t_{1} + t_{2} = 20 s$

Page No 197:

Answer:

Total moment of inertia of the system about the axis of rotation,
$I_{net} = (m_{1} r_{1}^{2} + m_{2} r_{2}^{2}) τ_{net} = F_{1} r_{1} - F_{2} r_{2} Also, τ_{net} = I_{net} \times α$
On equating the value of $τ_{n e t}$ and putting the value of I_net, we get:
$F_{1} r_{1} - F_{2} r_{2} = (m_{1} r_{1}^{2} + m_{2} r_{2}^{2}) \times α$
$(- 2 \times 10 \times 0.5) + (5 \times 10 \times 0.5) = [5 {(\frac{1}{2})}^{2} + 2 {(\frac{1}{2})}^{2}] α \Rightarrow 15 = \frac{7}{4} α \Rightarrow α = \frac{60}{7} = 8.57 rad / s^{2}$

Page No 197:

Answer:

Total moment of inertia of the system about the axis of rotation,
$I_{net} = (m_{1} r_{1}^{2} + m_{2} r_{2}^{2} + \frac{m l^{2}}{12}) m and l are the mass and length of the rod, respectively . τ_{net} = F_{1} r_{1} - F_{2} r_{2} Also, τ_{net} = I_{net} \times α$
On equating the value of $τ_{n e t}$ and putting the value of I_net, we get:
$F_{1} r_{1} - F_{2} r_{2} = (m_{1} r_{1}^{2} + m_{2} r_{2}^{2} + \frac{m l^{2}}{12}) \times α$
$(- 2 \times 10 \times 0.5) + (5 \times 10 \times 0.5) = [5 {(\frac{1}{2})}^{2} + 2 {(\frac{1}{2})}^{2} + \frac{{(1)}^{2}}{12}] α$
$\Rightarrow 15 = (1.75 + 0.084) α \Rightarrow α = \frac{1500}{(175 + 8.4)} = \frac{1500}{183.4} = 8.1 rad / s^{2} (g = 10) = 8.01 rad / s^{2} (if g = 9.8)$

(b) From the free body diagram of the block of mass 2 kg,
$T_{1} - m_{1} g = m_{1} a \Rightarrow T_{1} = 2 (a + g) = 2 (α r + g) (using, a = α r) = 2 (8 \times 0.5 + 9.8) \Rightarrow T_{1} = 27.6 N$

From the free body diagram of the block of mass 5 kg,
$m_{2} g - T_{2} = m_{2} a \Rightarrow T_{2} = m_{2} (g - a) = 5 (g - a) = 5 (9.8 - 8 \times 0.5) (a = α r) = 5 \times 5.8 = 29 N$

Page No 197:

Answer:

According to the equation, we have:

$M g - T_{1} = M a . . . (i) T_{2} = m a . . . (i i) \Rightarrow (T_{1} - T_{2}) = \frac{I a}{r^{2}} . . . (i i i) [Because, a = r a]$

If we add the equations (i) and (ii), we get:

$M g + T_{2} - T_{1} = M a + m a . . . (i v) \Rightarrow M g - I \frac{a}{r^{2}} = M a + m a \Rightarrow (M + m + \frac{I}{r^{2}}) a = M g \Rightarrow a = \frac{M g}{M + m + \frac{I}{r^{2}}}$

Page No 197:

Answer:

Moment of inertia of the bigger pulley, I = 0.20 kg-m²
r = 10 cm = 0.1 m,
Smaller pulley is light. Therefore, on neglecting its moment of inertia, we have:
Mass of the block, m = 2 kg

From the free body diagram, we get:
$m g - T = m a . . . (i)$
$T r = I α$ And $a = α r$
$\Rightarrow T = \frac{I a}{r^{2}} . . . (i i)$
Using equations (i) and (ii), we get:
$m g = (m + \frac{I}{r^{2}}) a \Rightarrow a = \frac{m g}{m + \frac{I}{r^{2}}} = \frac{2 \times 9.8}{2 + (\frac{0.2}{0.01})} = \frac{19.6}{22} = 0.89 m / s^{2}$

Page No 197:

Answer:

Given:
m = 2 kg, I₁ = 0.10 kg-m²
r₁ = 5 cm = 0.05 m
I₂ = 2.20 kg-m²
r₂ = 10 cm = 0.1 m

From the free body diagram, we have:
$m g - T_{1} = m a . . . (i) (T_{1} - T_{2}) r_{1} = I_{1} α . . . (i i) T_{2} r_{2} = I_{2} α . . . (i i i)$
Substituting the value of T₂ in the equation (ii), we get:
$\Rightarrow (T_{1} - I_{2} \frac{α}{r_{2}}) r_{1} = I_{1} α \Rightarrow T_{1} - I_{2} \frac{a}{r_{2}^{2}} = I_{1} \frac{a}{r_{1}^{2}} \Rightarrow T_{1} = \{(\frac{I_{1}}{r_{1}^{2}}) + (\frac{I_{2}}{r_{2}^{2}})\} a$
Substituting the value of T₁ in the equation (i), we get:
$m g - \{(\frac{I_{1}}{r_{1}^{2}}) + (\frac{I_{2}}{r_{2}^{2}})\} a = m a$
$\Rightarrow \frac{m g}{\{(\frac{I_{1}}{r_{1}^{2}}) + (\frac{I_{2}}{r_{2}^{2}})\} + m} = a \Rightarrow a = \frac{2 \times 9.8}{\frac{0.1}{0.0025} + \frac{0.2}{0.01} + 2} = 0.316 m / s^{2} \Rightarrow T_{2} = I_{2} \frac{a}{r_{2}^{2}} = \frac{0.20 \times 0.316}{0.01} = 6.32 N$

Page No 197:

Answer:

Free the body diagram of the system,

For block of mass M,
$M g - T_{1} = M a . . . (i)$
$(T_{1} - T_{2}) R = I α using, a = α r \Rightarrow (T_{1} - T_{2}) = I \frac{a}{R^{2}} . . . (i i) (For pully 1) Similarly, (T_{2} - T_{3}) = I \frac{a}{R^{2}} . . . (i i i) (For pully 2)$

For block of mass m,
$T_{3} - m g = m a . . . (i v) (For block m)$

Adding equations (ii) and (iii), we get:
$(T_{1} - T_{3}) = \frac{2 I a}{R^{2}} . . . (v)$
Adding equations (i) and (iv), we get:
$- m g + M g + (T_{3} - T_{1}) = M a + m a . . . (v i)$
Using equations (v) and (vi), we get:
$M g - m g = M a + m a + \frac{2 I a}{R^{2}} \Rightarrow a = \frac{(M - m) g}{(M + m + \frac{2 I}{R^{2}})}$

Page No 197:

Answer:

Let the mass of block be m₁ and mass of pulley be m₂.
Acceleration of the massive pulley will be half of that of the block.

From the free body diagram, we have:
$T_{1} = m_{1} a . . . (i) (T_{2} - T_{1}) r = I α T_{2} - T_{1} = \frac{I a}{2 r^{2}} = \frac{5 a}{2} . . . (i i) m_{2} g - m_{2} \frac{a}{2} = T_{1} + T_{2} . . . (i i i)$
Putting the value of mass in equation (i) and using equation (i) in equation (ii), we get:
$T_{1} = a and T_{2} = \frac{7}{2} a$
$m_{2} g = m_{2} \frac{a}{2} + \frac{7}{2} a + a On replacing the value of m_{2} using \frac{1}{2} m r^{2} = I, we get : \frac{2 I}{r^{2}} g = \frac{2 I}{r^{2}} \frac{a}{2} + \frac{9}{2} a \Rightarrow 98 = 5 a + 4.5 a \Rightarrow a = \frac{98}{9.5} = 10.3 m / s^{2}$

Page No 197:

Answer:

From the figure, we have:
$T_{1} - m_{1} g \sin θ = m_{1} a . . . (i) T_{2} - T_{1} = I \frac{a}{r^{2}} . . . (i i) m_{2} g \sin θ - T_{2} = m_{2} a . . . (i i i)$
Adding equations (i) and (iii), we get:
$m_{2} g \sin θ + (- T_{2} + T_{1}) - m_{1} g \sin θ = (m_{1} + m_{2}) a$

$\Rightarrow (m_{2} - m_{1}) g \sin θ + (- \frac{I}{r^{2}}) a = (m_{1} + m_{2}) a \Rightarrow (m_{2} - m_{1}) g \sin θ = (m_{1} + m_{2}) a + (\frac{I}{r^{2}}) a \Rightarrow a = \frac{(m_{2} - m_{1}) g \sin θ}{(m_{2} + m_{1} + \frac{I}{r^{2}})} = \frac{(4 - 2) \times 10 \times (1 \sqrt{2})}{(4 + 2) + (0.5 / 0.01)} = \frac{(2 \times 10 \times \frac{1}{\sqrt{2}})}{(6 + 50)} = 0.248 = 0.25 m / s^{2}$

Page No 197:

Answer:

From the figure, we have:
$m_{2} g s in θ - T_{2} = m_{1} a . . . (i) T_{1} - (m_{1} g \sin θ + μ m_{1} g \cos θ) = m_{1} a . . . (i i) T_{2} - T_{1} = \frac{I a}{r^{2}} . . . (i i i)$
Adding equations (i) and (ii), we get:
$m_{2} g \sin θ - (m_{1} g \sin θ - μ m_{1} g \cos θ) + (T_{1} - T_{2}) = m_{1} a + m_{2} a$
$m_{2} g \sin θ - (m_{1} g \sin θ + μ m_{1} g \cos θ) - \frac{I a}{r^{2}} = m_{1} a + m_{2} a m_{2} g \sin θ - (m_{1} g \sin θ + μ m_{1} g \cos θ) = m_{1} a + m_{2} a + \frac{I a}{r^{2}}$
$4 \times 9.8 \times \frac{1}{\sqrt{2}} - \{2 \times 9.8 \times \frac{1}{\sqrt{2}} + 0.5 \times 2 \times 9.8 \times \frac{1}{\sqrt{2}}\} = (4 + 2 + \frac{0.5}{0.01}) a \Rightarrow 27.80 - 13.90 + 6.95 = 56 a \Rightarrow 27.8 - 20.8 = 56 a \Rightarrow a = \frac{7}{56} = 0.125 m / s^{2}$

Page No 197:

Answer:

Given:
Mass of the stick = $m_{1} = 200 g$
Mass of the small object = $m_{2} = 20$
Length of the string = $l = 1 m$
As the system is in equilibrium, we have:
$τ_{t o t a l} = 0 (about O)$
$(T_{1} \times r_{1}) - (T_{2} \times r_{2}) - (m_{1} g \times r_{3}) = 0$

$\Rightarrow T_{1} \times 0.7 - T_{2} \times 0.3 - 2 \times 0.2 \times g = 0 \Rightarrow 7 T_{1} - 3 T_{2} = 3.92 . . . (i)$
Now, we have:
Total upward force = Total downward force
$T_{1} + T_{2} = m_{1} g + m_{2} g = 0.2 \times 9.8 + 0.02 \times 9.8 \Rightarrow T_{1} + T_{2} = 2.156 . . . (i i)$

Solving equations (i) and (ii), we get:
$T_{1} = 1.038 N \approx 1.04 N; T_{2} = 1.18 \approx 1.12 N$

Page No 197:

Answer:

Let N₂ be the normal force on the ladder by the ground.
Let N₁ be the normal force on the ladder by the wall.
Let f_r be the force of friction on the ladder by the ground.
As system is in translation equilibrium, we have:
$N_{1} = f_{r} = μ N_{2}; N_{2} = 16 g + 60 g = 745 N$
Applying condition of rotational equilibrium at point O,
i.e., about point O, we have:
$Γ_{t o t a l} = 0$
$N_{1} \times 10 \cos 37 ° = 16 g \times 5 \sin 37 ° + 60 g \times 8 \sin 37 °$
$\Rightarrow 8 N_{1} = 48 g + 288 g \Rightarrow N_{1} = \frac{336 \times 9.8}{8} = 412 N ∴ μ = \frac{N_{1}}{N_{2}} = \frac{412}{745} = 0.553$

Page No 198:

Answer:

Let the maximum mass of a mechanic who could go up be m.
The system is in translation and rotational equilibrium; therefore, we have:
$N_{2} = 16 g + m g . . . (i) N_{1} = μ N_{2} . . . (i i) R_{1} \times 10 \cos 37 ° = 16 g \times 5 \sin 37 ° - m g \times 8 \sin 37 ° . . . (i i i)$

$\Rightarrow 8 N_{1} = 48 g + \frac{24}{5} m g From eq . (i i), we have : N_{2} = \frac{48 g + \frac{24}{5} mg}{8 \times 0.54}$

Putting the value of N₂ in eq. (i), we have:
$16 g + m g = \frac{24.0 g + 24 m g}{5 \times 8 \times 0.54} \Rightarrow 16 + m = \frac{24.0 + 24 m}{40 \times 0.54} \Rightarrow m = 44 kg$

Therefore, weight of the mechanic, who can go up and do the work, should be less than 44 kg.

Page No 198:

Answer:

Given:
Mass of the man = m = 60 kg
Ladder length = 6.5 m
Height of the wall = 6 m

(a) We have to find the torque due to the weight of the body about the upper end of the ladder.
$τ = 60 \times 10 \times \frac{6.5}{2} \sin θ \Rightarrow τ = 600 \times \frac{6.5}{2} \times \sqrt{(1 - \cos^{2} θ)} \Rightarrow τ = 600 \times (\frac{6.5}{2}) \times \sqrt{\{1 - {(\frac{6}{6.5})}^{2}\}} \Rightarrow τ = 740 N - m$

(b) Let us find the vertical force exerted by the ground on the ladder.
$R_{2} = m g = 60 \times 9.8 = 588 N$
Vertical force exerted by the ground on the ladder = $μ R_{2} = R_{1}$
As system is in rotational equilibrium, we have:
$τ_{net} = 0 (about O)$
$\Rightarrow 6.5 N_{1} \cos θ = 60 g \times \frac{6.5}{2} \sin θ$
$\Rightarrow N_{1} = \frac{1}{2} 60 g \tan θ = \frac{1}{2} 60 g \times (\frac{2.5}{6}) [using, \tan θ = \frac{2.5}{6}] \Rightarrow N_{1} = \frac{25}{2} g \Rightarrow N_{1} = 122.5 N \approx 120 N$

Page No 198:

Answer:

It is given that the magnitudes of the forces exerted by the hinges on the door are equal.

Therefore, we have:
Resultant of N₁ and F₁ at point A = Resultant of N₂ and F₂ at point B
$\Rightarrow \sqrt{{(N_{1})}^{2} + {(F_{1})}^{2}} = \sqrt{{(N_{2})}^{2} + {(F_{2})}^{2}}$ ...(i)
System is in translation equilibrium. Therefore, we have:
N₁ = N₂;
$8 g = F_{1} + F_{2}$
Putting the value in eq. (i), we get:
F₁ = F₂
$\Rightarrow 2 F_{1} = 8 g \Rightarrow F_{1} = 40$
Let us take the torque about point B. We get:
$N_{1} \times 4 = 8 g \times 0.75 \Rightarrow N_{1} = \frac{(80 \times 3)}{4 \times 4} = 15 N$
Now, the forces exerted by the hinges A on the door,
$\sqrt{(F_{1}^{2} + N_{1}^{2})} = \sqrt{(40^{2} + 15^{2})} = 42.72 = 43 N$

Page No 198:

Answer:

Given:
Length of rod = L
It makes an angle θ with the floor.
Height of the wall is h.

The system is in translational equilibrium; therefore, we get:

$N_{2} = m g - N_{1} \cos θ . . . (i) N_{1} \sin θ = μ N_{2} . . . (i i) System is in rotational equilibrium; therefore, we have : N_{1} \cos θ \times \frac{h}{\tan θ} + N_{1} \sin θ \times h = m g \times \frac{L}{2} \cos θ \Rightarrow N_{1} = \frac{m g \times \frac{L}{2} \cos θ}{(\frac{\cos^{2} θ}{\sin θ}) h + h \sin θ} \Rightarrow N_{1} cosθ = \frac{\frac{1}{2} m g L \cos^{2} θ}{\{(\frac{\cos^{2} θ}{\sin θ}) h + h \sin θ\}}$
From equation (ii), we have:
$μ = \frac{N_{1} \sin θ}{N_{2}} = \frac{N_{1} \sin θ}{m g - N_{1} \cos θ} \Rightarrow μ = \frac{L / 2 \cos θ \sin θ \times 2 \sin θ}{2 (\cos^{2} θ + \sin^{2} θ) h - L \cos^{2} θ \sin θ} \Rightarrow μ = \frac{L \cos θ \sin^{2} θ}{2 h - L \cos^{2} θ \sin θ}$

Page No 198:

Answer:

Given:
Mass of the rod = m = 300 gm
Length of the rod = l = 50 cm
Angular velocity of the rod = $ω =$ 2 rad/s

(a) Moment of inertia about one end of the rod,
$I = \frac{m L^{2}}{3} = \frac{\{0.3 \times {(0.50)}^{2}\}}{3} \Rightarrow I = \frac{0.25}{10} = 0.025 kg - m^{2}$
Angular momentum about that point, $L = I ω$
L = 0.025 × 2 = 0.05 kg-m²/s

(b) Speed of the centre of the rod,
$v = ω r = 2 \times \frac{0.50}{2} = 0.5 m / s$

(c) Kinetic energy, $K = \frac{1}{2} I ω^{2}$
$\Rightarrow K = \frac{1}{2} \times 0.025 \times 2^{2} = \frac{1}{2} \times 0.025 \times 4 = 0.05 joule$

Page No 198:

Answer:

Given:
Torque acting on the plate = $τ = 0.10 N - m$
Mass of the plate = $m = 2 kg$
$On applying τ = I α, we get : \frac{m r^{2}}{12} \times α = 0.10 N - m \Rightarrow α = 60 rad / s$

Let ω be the angular velocity after time t (t = 5 s).
$Therefore, we have : ω = ω_{0} + a t \Rightarrow ω = 60 \times 5 = 300 rad / s Angular momentum, I ω = (\frac{0.10}{60}) \times 300 = 0.50 kg - m^{2} / s Kinetic energy, \frac{1}{2} I ω^{2} = \frac{1}{2} \times (\frac{0.10}{60}) \times 300^{2} = 75 joule$

Page No 198:

Answer:

Given:
r = 6400 km = 6.4 × 10⁶ m;
R = 1⋅5 × 10⁸ km = 1⋅5 × 10¹¹ m
About its axis, we have:
T = 1 day = 86400 s;
$ω = \frac{2 π}{T}$
Angular momentum of the Earth about its axis,
$L = I ω = \frac{2}{5} m r^{2} \times (\frac{2 π}{86400})$

About the Sun's axis,
T = 365 day = 365 × 86400 s
Angular momentum of the Earth about the Sun's axis,
$L' = m R^{2} \times (\frac{2 π}{86400 \times 365})$
Ratio of angular momentums,

$\frac{L}{L'} = \frac{2 / 5 m r^{2} \times (2 π / 86400)}{m R^{2} \times 2 π / (86400 \times 365)} = \frac{(2 r^{2} \times 365)}{5 R^{2}} = (\frac{2 \times {(6.4 \times 10^{6})}^{2} \times 365}{5 \times {(1.5 \times 10^{11})}^{2}}) = 2.65 \times 10^{- 7}$

Page No 198:

Answer:

Distance of centre of mass from mass m₁,
$d_{1} = \frac{m_{2} r}{m_{1} + m_{2}}$
Angular momentum due to the mass m₁ at the centre of system,
$L_{1} = I_{1} ω = m_{1} {d_{1}}^{2} ω = m_{1} {(\frac{m_{2} r}{m_{1} + m_{2}})}^{2} ω$

$\Rightarrow L_{1} = \frac{m_{1} m_{2}^{2} r^{2}}{{(m_{1} + m_{2})}^{2}} ω . . . (1)$

Similarly, the angular momentum due to the mass m₂ at the centre of system,
$L_{2} = m_{2} {(\frac{m_{1} r}{m_{1} m_{2}})}^{2} ω = \frac{m_{2} m_{1}^{2} r^{2}}{{(m_{1} + m_{2})}^{2}} ω . . . (2)$

Therefore, net angular momentum,

$L = L_{1} + L_{2} = \frac{m_{1} m_{2}^{2} r^{2} ω}{{(m_{1} + m_{2})}^{2}} + \frac{m_{2} m_{1}^{2} r^{2} ω}{{(m_{1} + m_{2})}^{2}} = \frac{m_{1} m_{2} (m_{1} + m_{2}) r^{2} ω}{{(m_{1} + m_{2})}^{2}} = \frac{m_{1} m_{2}}{{(m_{1} + m_{2})}^{2}} r^{2} ω = μ r^{2} ω$

Page No 198:

Answer:

Moment of inertia of the dumb-bell,
$I = m r^{2} + m r^{2} = 2 m r^{2}$
Torque, $τ = I α$
$\Rightarrow F \times r = (m r^{2} + m r^{2}) α \Rightarrow 5 \times 0.25 = 2 m r^{2} \times α \Rightarrow α = \frac{1.25}{2 \times 0.5 \times {(0.25)}^{2}} = 20 rad / s^{2}$

Given:
$ω_{0} = 10 rad / s and t = 0.10 s$
$Using ω = ω_{0} + α t, we get : ω = 10 + 20 \times 0.10 = 10 + 2 = 12 rad / s$

Page No 198:

Answer:

Given:
Initial moment of inertia of the system,
I₁ = 0.500 kg-m²;
r = 0.2 m;
ω = 20 rad/s
Mass of the stationary particle, m = 0.2 kg
Final moment of inertia of the system,
I₂ = I₁+ mr²
It is given:
External torque = 0
Angular momentum is conserved; therefore, we have:
$I_{1} ω_{1} = I_{2} ω_{2}$
$\Rightarrow 0.5 \times 20 = (0.5 + 0.2 \times {(0.2)}^{2}) ω_{2} \Rightarrow ω_{2} = \frac{10}{0.508} \approx 19.7 rad / s$

Page No 198:

Answer:

Initial moment of inertia of diver, I₁ = 6 kg-m²
Initial angular velocity of diver, ω₁ = 2 rad/s
Final moment of inertia of diver, I₂ = 5 kg-m²
Let ω₂ be the final angular velocity of the diver.
We have:
External torque = 0
$∴$ I₁ω₁ = I₂ω₂
$\Rightarrow ω_{2} = \frac{6 \times 2}{5} = 2.4 rad / s$

Page No 198:

Answer:

Given:
Initial angular speed of the system,
$ω_{1} = 120 rpm = 120 \times (\frac{2 π}{60}) = 4 π rad / s$
Initial moment of inertia of the system,
$I_{1} = 6 kg - m^{2}$
Final moment of inertia of the system,
$I_{2} = 2 kg - m^{2}$
Two balls are inside the system; therefore, we get:
Total external torque = 0
$∴$ $I_{1} ω_{1} = I_{2} ω_{2}$

$\Rightarrow 6 \times 4 π = 2 ω_{2} \Rightarrow ω_{2} = 12 π rad / s Or, ω_{2} = 6 rev / s = 360 rev / minute$

Page No 198:

Answer:

Given:
Moment of inertia of umbrella = I₁ = 2 × 10⁻³ kg-m²
Moment of inertia of the system = I₂ = 3 × 10⁻³ kg-m²
Angular speed of the umbrella with respect to the boy = ω₁ = 2 rev/s
Let the angular velocity imparted to the platform be ω₂.
Taking the Earth as the reference, we have:
Angular velocity of the umbrella = (ω₁ − ω₂)
Applying conservation of angular momentum, we get:
$I_{1} (ω_{1} - ω_{2}) = I_{2} (ω_{2}) \Rightarrow 2 \times 10^{- 3} (2 - ω_{2}) = 3 \times 10^{- 3} ω_{2} \Rightarrow 5 ω_{2} = 4 \Rightarrow ω_{2} = 0.8 rev / s$

Page No 198:

Answer:

Given:
For the first wheel,
I₁ = 10 kg-m² and ω₁ = 160 rev/min
For the second wheel, ω₂ = 300 rev/min
Let I₂ be the moment of inertia of the second wheel.
After they are coupled, we have:
ω = 200 rev/min
If we take the two wheels to be an isolated system, we get:
Total external torque = 0
Therefore, we have:
$I_{1} ω_{1} + I_{2} ω_{2} = (I_{1} + I_{2}) ω$
$\Rightarrow 0.10 \times 160 + I_{2} \times 300 = (0.10 + I_{2}) \times 200 \Rightarrow 16 + 300 I_{2} = 20 + 200 I_{2} \Rightarrow 100 I_{2} = 4 \Rightarrow I_{2} = \frac{4}{100} = 0.04 kg - m^{2}$

Page No 198:

Answer:

On considering two bodies as a system, we get:
Moment of inertia of kid and ball about the axis $= (M + m) R^{2}$
Applying the law of conservation of angular momentum, we have:
$m ν R = \{I + (M + m) R^{2}\} ω$
$\Rightarrow ω = \frac{m ν R}{I + (M + m) R^{2}}$

Page No 198:

Answer:

Initial angular momentum of the system is zero (L₁ = 0).
Let the platform starts rotating with an angular velocity ω.
Therefore, we have:
Final angular momentum of the system = L₂ $= I_{ω} + M R^{2} ω - m v R$
(m = mass of the ball, v = velocity of the ball)
Total external torque = 0
Also,
Initial angular momentum = Final angular momentum
$\Rightarrow I ω + M R^{2} ω - m v R = 0 \Rightarrow ω = \frac{m v R}{(I + M R^{2})}$

Page No 198:

Answer:

Let ω' be the angular velocity of platform after the kid starts walking.
Let ω be the angular velocity of the wheel before walking.
When we see the (kid-wheel) system from the initial frame of reference, we can find that the wheel moves with a speed of $ω$ and the kid with a speed of $(ω' + \frac{ν}{R})$ , after the kid has started walking.
Initial angular momentum of the system = (I + MR²)ω
Angular velocity of the kid after it starts walking = $(ω' + \frac{v}{R})$
External torque is zero; therefore, angular momentum is conserved.
$\Rightarrow (I + M R^{2}) ω = I ω' + M R^{2} (ω' + \frac{v}{R}) \Rightarrow (I + M R^{2}) ω = I ω' + M R^{2} (ω' + \frac{v}{R}) \Rightarrow (I + M R^{2}) ω - M v R = (I + M R^{2}) ω' \Rightarrow ω' = \frac{(I + M R^{2}) ω - M v R}{(I + M R^{2})} \Rightarrow ω' = ω - \frac{M v R}{I + M R^{2}}$

Page No 199:

Answer:

Given:
Mass of the rod = m
Length of the rod = l

(a) For the centre of mass,
Acceleration, $a = \frac{F}{m}$
$Velocity, v = a t \Rightarrow v = \frac{F t}{m}$

(b) Let the angular speed about the centre of mass be ω.
Moment of inertia of rod about centre of mass = $I = \frac{m l^{2}}{12}$
$I ω = m v r$
$\Rightarrow \frac{m l^{2}}{12} \times ω = m ν \frac{l}{2} \Rightarrow \frac{m l^{2}}{12} \times ω = m \times \frac{F t}{m} \times \frac{l}{2} \Rightarrow ω = 6 \frac{F t}{m l}$

(c) Kinetic energy,
$K . E . = \frac{1}{2} m ν^{2} + \frac{1}{2} I ω^{2} = \frac{1}{2} \times m {(\frac{F t}{m})}^{2} + \frac{1}{2} \times I ω^{2} = \frac{1}{2} \times m (\frac{F^{2} t^{2}}{m^{2}}) + \frac{1}{2} \times \frac{m l^{2}}{12} \times (\frac{36 F^{2} t^{2}}{m^{2} l^{2}}) = \frac{2 F^{2} t^{2}}{m}$

(d) Angular momentum about the centre of mass,
$L = m ν r = m \times \frac{F t}{m} \times (\frac{l}{2}) = \frac{F l t}{2}$

Page No 199:

Answer:

Let initial velocity of the particle = u₁
Let final velocity of the particle = v₁
Let time taken to move π/2 angle = t
Given v₁= 0
Let initial velocity of CM of the rod = u₂
Given u₂= 0
Let final velocity of the CM of rod = v₂
By conservation of linear momentum
mu₁ + Mu₂ = mv₁ + Mv₂
=> v₂ = (m/M)u₁ ---(1)
Now by we consider the angular momentum imparted by the particle to the rod.
mu₁(L/2) = Iω
For rod about its CM, I = ML₂/12
mu₁(L/2) = ωML₂/12
=> ω = 6mu1/ML ---2.
Now we know
ω = θ/t
θ = π/2
=> t = θ/ω = (π/2)/(6mu₁/ML)
=> t = πML/(12mu₁)
Linear distance moved by the CM of the rod will be
s = v₂t
By equation 1.
s = [πML/(12mu₁)]× (m/M)u₁
= πL/12
Initial KE of the particle is ½m(u₁)²
KE of the CM of the rod = ½M(v₁)² = ½M{(m/M)u₁}² [by equation 1.]
= ½(u₁)²m₂/M
KE_rot of the rod = ½Iω²
KE_rot = ½ (ML₂/12)(6mu₁/ML)²
= (3/2) m²(u₁)²/M
To be elastic collision KE must be conserved
½m(u₁)² = ½(u₁)²m^₂/M + (3/2) m²(u₁)²/M
=> M = 4m

Hence Proved.

Page No 199:

Answer:

(a) It is given that no external torque and force is applied on the system.
Applying the law of conservation of momentum, we get:
$m ν = (M + m) ν'$
$\Rightarrow ν' = \frac{m ν}{M + m}$

(b) Velocity of the particle w.r.t. centre of mass (COM) C before the collision = $v_{c} = v - v'$
$\Rightarrow v_{c} = v - \frac{m v}{M + m} = \frac{M v}{M + v}$

(c) Velocity of the particle w.r.t. COM C before collision $= - \frac{M ν}{M + m}$

(d) Distance of the COM from the particle,
$x_{c m} = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}} \Rightarrow r = \frac{M \times \frac{L}{2} + m \times 0}{M + m} \Rightarrow r = \frac{M L}{2 (M + m)}$
∴ Angular momentum of body about COM

$= m v r = m \times \frac{M v}{(M + m)} \times \frac{M L}{2 (M + m)} = \frac{M^{2} m v L}{2 {(M + m)}^{2}}$

∴ Angular momentum of rod about COM

$= M \times (\frac{m v}{(M + m)}) \times \frac{1}{2} \frac{m L}{(M + m)} = \frac{M m^{2} v L}{2 {(M + m)}^{2}}$

(e) Moment of inertia about COM = I $= I_{1} + I_{2}$

$I = m {[\frac{M L}{2 (M + M)}]}^{2} + \frac{M L^{2}}{12} + M {[\frac{M L}{2 (m + M)}]}^{2} = \frac{m M^{2} L^{2}}{4 {(m + M)}^{2}} + \frac{M L^{2}}{12} + \frac{M m^{2} L^{2}}{4 {(M + m)}^{2}} = \frac{3 m M^{2} L^{2} + M {(m + M)}^{2} L^{2} + 3 M m^{2} L^{2}}{12 {(m + M)}^{2}} = \frac{M (M + 4 m) L^{2}}{12 (M + m)}$

(f) About COM,
$V_{c m} = \frac{m ν}{M + m} ∴ I ω = m v r = m v \times \frac{M L}{2 (M + m)} \Rightarrow ω = \frac{m ν M L}{2 (M + m)} \times \frac{12 (M + m)}{M (M + 4 m) L^{2}} = \frac{6 m ν}{(M + 4 m) L}$

Page No 199:

Answer:

Let the new angular speed of the rod be ω'.
Here, net torque on the system is zero.
So, the angular momentum is conserved.
Therefore, we get:
$I ω = I' ω'$
$[m {(\frac{L}{2})}^{2} + m {(\frac{L}{2})}^{2}] ω = [2 m {(\frac{L}{2})}^{2} + m {(\frac{L}{2})}^{2}] ω'$
$2 m L^{2} ω = 3 m L^{2} ω' \Rightarrow ω' = \frac{2}{3} ω$

Page No 199:

Answer:

(a) Collision will not affect the velocity of ball B because the light rod will exert a force on it only along its length.
Therefore, we have:
Velocity of B = v₀

For ball A,

On applying the law of conservation of linear momentum, we get:
$m ν_{0} = 2 m \times ν$
$\Rightarrow v = \frac{v_{0}}{2}$
$∴$ Velocity of A $= \frac{ν_{0}}{2}$

(b) If we consider the three bodies to be a system, we have:
Net external force = 0
$v_{CM} = \frac{m \times v_{0} + 2 m \times (v_{0} / 2)}{n + 2 m} = \frac{m v_{0} + m v_{0}}{3 m} = \frac{2 v_{0}}{3}$
(Direction will be same as the initial velocity before collision.)

(c) Velocity of (A + P) w.r.t. the centre of mass $= \{\frac{2 v_{0}}{3} - \frac{v_{0}}{2}\} = \frac{v_{0}}{6}$
Velocity of B w.r.t. the centre of mass $= v_{0} - \frac{2 v_{0}}{3} = \frac{v_{0}}{3}$
Distance of the (A + P) from the centre of mass $= \frac{l}{3}$
Distance of the B from the centre of mass = $\frac{2 l}{3}$
$Applying m v_{c o m} r = l_{c m} \times ω, we have : (2 m \times \frac{v_{o}}{6} \times \frac{l}{3}) + (m \times \frac{v_{0}}{3} \times \frac{2 l}{3}) = \{2 m {(\frac{l}{3})}^{2} + {(\frac{2 l}{3})}^{2} m\} \times ω \Rightarrow \frac{6 m v_{0} l}{18} = \frac{6 m l}{9} ω \Rightarrow ω = \frac{v_{0}}{2 l}$

Page No 199:

Answer:

(a) Angular momentum $= m ν r$
Let the particle P collides the ball B with a speed u and system moves with speed v just after the collision.
Applying the law of conservation of linear momentum, we get:
$m u = 2 m ν - m ν = m ν$
$∴ u = ν$
$Velocity, u = \sqrt{2 g h} and r = \frac{L}{2}$

Initial angular momentum of system about COM of rod,
$m u r = m \times \sqrt{2 g h} \times \frac{L}{2} = \frac{m L \sqrt{g h}}{\sqrt{2}}$
Angular momentum of system about COM of the rod just after the collision $= l ω$
$I = \frac{2 m L^{2}}{4} + \frac{m L^{2}}{4} = \frac{3 m L^{2}}{4}$
Applying the law of conservation of angular momentum and obtaining the value of ω, we get:
$ω = \frac{m L \frac{\sqrt{g h}}{\sqrt{2}}}{\frac{3 m L^{2}}{4}} = \frac{\sqrt{8 g h}}{3 L}$

(b) When the mass 2m and m are at the top most position and at the lowest point, respectively, they will automatically rotate. In this position, we have:
Total gain in potential energy $= 2 m g (\frac{L}{2}) - m g (\frac{L}{2})$
Kinetic energy = $\frac{1}{2} I ω^{2}$
Therefore, by the law of conservation of energy, we have:
$m g \frac{L}{2} = \frac{1}{2} \times \frac{3}{4} m L^{2} \times (\frac{8 g h}{9 g L^{2}}) \Rightarrow h = \frac{3 L}{2}$

Page No 199:

Answer:

From the free body diagram, we have:
$0.4 g - T_{1} = 0.4 a . . . (i) T_{2} - 0.2 g = 0.2 a . . . (i i) (T_{1} - T_{2}) r = \frac{l a}{r} . . . (i i i)$

On solving the above equations, we get:
$a = \frac{(0.4 - 0.2) g}{(0.4 + 0.2 + \frac{1.6}{0.4})} = \frac{g}{5}$
On solving the (b) part of the question first, we have:
Speed of the blocks = $v = \sqrt{(2 a h)} = \sqrt{2 \times \frac{g}{5} \times 0.5} = \sqrt{(\frac{9.8}{5})} = 1.4 m / s$

(a) Total kinetic energy of the system
$= \frac{1}{2} m_{1} v^{2} + \frac{1}{2} m_{2} v^{2} + \frac{1}{2} I ω^{2} = \frac{1}{2} m_{1} v^{2} + \frac{1}{2} m_{2} v^{2} + \frac{1}{2} I {(\frac{v}{r})}^{2} = (\frac{1}{2} \times 0.4 \times {(1.4)}^{2}) + (\frac{1}{2} \times 0.2 \times {(1.4)}^{2}) + (\frac{1}{2} \times 1.6 \times 10^{- 4} \times {(\frac{1.4}{2 \times 10^{- 2}})}^{2}) = (0.2 + 0.1 + 0.2) {(1.4)}^{2} = 0.5 \times 1.96 = 0.98 joule$

Page No 199:

Answer:

Given:
Moment of inertia of the pully = I = 0.2 kg-m²
Radius of the pully = r = 0.2 m
Spring constant of the spring = k = 50 N/m
Mass of the block = m = 1 kg
g = 10 ms² and h = 0.1 m
On applying the law of conservation of energy, we get:
$m g h = \frac{1}{2} m v^{2} + \frac{1}{2} k x^{2} + \frac{1}{2} I {(\frac{ω}{r})}^{2}$
On putting x = h = 0.1 m, we get:
$1 = \frac{1}{2} \times 1 \times v^{2} + \frac{1}{2} \times 0.2 \times \frac{v^{2}}{0.04} + \frac{1}{2} \times 50 \times 0.01 \Rightarrow 1 = 0.5 v^{2} + 2.5 v^{2} + \frac{1}{4} \Rightarrow 3 v^{2} = \frac{3}{4} \Rightarrow v^{2} = \frac{1}{4} \Rightarrow v = \frac{1}{2} = 0.5 m / s$

Page No 199:

Answer:

Let the mass of the rod and its angular velocity be m and ω (when hits the ground), respectively.
It is given that the rod has rotational motion only.

On applying the law of conservation of energy, we get:
$\frac{1}{2} I ω^{2} = m g \frac{l}{2} \Rightarrow \frac{m l^{2}}{3} \times ω^{2} = m g l \Rightarrow ω^{2} = \frac{3 g}{l} \Rightarrow ω = \sqrt{\frac{3 g}{l}} = \sqrt{(3 \times \frac{9.8}{1})} \Rightarrow ω = 5.42 rad / s$

Page No 199:

Answer:

Let the angular velocity of stick after the collision be ω.
Given:
Mass of the stick = m = 240 g
Mass of the particle = m' = 100 g
Length of the string (or rod) = r = l = 1 m
Moment of inertia of the particle about the pivoted end = $I' = m' r^{2}$
$\Rightarrow I' = (0.1 \times 1^{2}) \Rightarrow I' = 0.1 kg - m^{2}$
Moment of inertia of the rod about the pivoted end = $I = \frac{m l^{2}}{3}$
$\Rightarrow I = (\frac{0.24}{3} \times 1^{2}) \Rightarrow I = 0.08 kg - m^{2}$

Applying the law of conservation of energy, we get:
Final energy of the particle = Initial energy of the rod
$m g h = \frac{1}{2} I ω^{2} \Rightarrow \frac{1}{2} I ω^{2} = 0.1 \times 10 \times 1 \Rightarrow ω = \sqrt{20}$

For collision, we have:
$0.1 \times 1^{2} \times \sqrt{20} + 0$
$= [(\frac{0.24}{3}) \times 1^{2} + {(0.1)}^{2} {(1)}^{2}] ω \Rightarrow ω = \frac{\sqrt{20}}{[10 \times (0.18)]} \Rightarrow 0 - \frac{1}{2} I ω^{2} = - m_{1} g l (1 - \cos θ) - m_{2} g \frac{1}{2} (1 - \cos θ) \Rightarrow \frac{1}{2} I ω^{2} = - m g l (1 - \cos θ) - m_{2} g \frac{1}{2} (1 - \cos θ)$
$\Rightarrow \frac{1}{2} I ω^{2} = 0.1 \times 10 (1 - \cos θ) - 0.24 \times 10 \times 0.5 (1 - \cos θ) \Rightarrow \frac{1}{2} \times 0.18 \times (\frac{20}{324}) = 2.2 \times 9 (1 - \cos θ) \Rightarrow (1 - \cos θ) = \frac{1}{(2.2 \times 1.8)} \Rightarrow 1 - \cos θ = 0.252 \Rightarrow \cos θ = 1 - 0.252 = 0.748 \Rightarrow θ = \cos^{- 1} (0.748) = 41 °$

Page No 199:

Answer:

Let the length of the rod be l.
Mass of the rod be m.
Let the angular velocity of the rod be ω when it makes an angle of 37° with the vertical.

On applying the law of conservation of energy, we get:

$\frac{1}{2} I ω^{2} - 0 = m g \frac{l}{2} (\cos 37 ° - \cos 60 °) \Rightarrow \frac{1}{2} \times \frac{m l^{2} ω^{2}}{3} = m g \frac{l}{2} (\frac{4}{5} - \frac{1}{2}) \Rightarrow ω^{2} = \frac{9 g}{10 l}$

Let the angular acceleration of the rod be α when it makes an angle of 37° with the vertical.
$Using τ = I α, we get : I α = m g \frac{l}{2} \sin 37 ° \Rightarrow \frac{m l^{2}}{3} α = m g \frac{l}{2} \times \frac{3}{5} \Rightarrow α = 0.9 (\frac{g}{l})$

Force on the particle of mass dm at the tip of the rod
$F_{c} = centrifugal force = (d m) ω^{2} l = (d m) \frac{9 g}{10 l} l \Rightarrow F_{c} = 0.9 g (d m) F_{t} = tangential force = (d m) α l \Rightarrow F_{t} = 0.9 g (d m)$

So, total force on the particle of mass dm at the tip of the rod will be the resultant of F_cand F_t.

$∴ F = \sqrt{({F_{c}}^{2} + {F_{t}}^{2})} = 0.9 \sqrt{2} g (d m)$

Page No 199:

Answer:

Let v_c be the translational velocity of the cylinder.
Let ω be the rotational velocity of the cylinder.
Let r be the radius of the cylinder.
For rolling, we have:
v_c= rω

Speed of the highest point = v_c+ rω = 2v_c
⇒ $2 \times 25 m / s$ = 50 m/s

Page No 200:

Answer:

Let radius of the sphere be R and its angular speed be ω.
Moment of inertia of sphere, $I = \frac{2}{5} m R^{2}$
Total kinetic energy,
$K = \frac{1}{2} I ω^{2} + \frac{1}{2} m v^{2} K = \frac{1}{2} \times (\frac{2}{5} m R^{2}) \times \frac{v^{2}}{R^{2}} + (\frac{1}{2} m v^{2}) K = \frac{2}{10} m v^{2} + \frac{1}{2} m v^{2} K = \frac{(2 + 5) m v^{2}}{10} = \frac{7}{10} m v^{2}$

Page No 200:

Answer:

Let the radius of the disc be R.
Let the tension in the string be T.
Let the acceleration of the disc be a.

From the free body diagram, we have:
$m g - T = m a$ ...(i)
Torque about the centre of disc,
$T \times R = I \times α$
$\Rightarrow T \times R = \frac{1}{2} m R^{2} \times \frac{a}{R} \Rightarrow T = \frac{1}{2} m a . . . (ii)$

Putting this value in equation (i), we get:
$m g - \frac{1}{2} m a = m a \Rightarrow m g = \frac{3}{2} m a \Rightarrow a = \frac{2 g}{3}$

Page No 200:

Answer:

Let r be the radius of the ball.
Let v be the linear speed of the ball when it rolls on the horizontal part of the track.
Let ω be the angular speed of the ball when it rolls on the horizontal part of the track.
Potential energy the ball has gained w.r.t. the surface will be converted to angular kinetic energy about the centre and linear kinetic energy.
Therefore, we have:
$m g h = \frac{1}{2} I ω^{2} + m v^{2}$
$\Rightarrow m g h = \frac{1}{2} \times (\frac{2}{5} m R^{2}) \times {(\frac{v}{R})}^{2} + \frac{1}{2} m v^{2} \Rightarrow g h = \frac{1}{5} v^{2} + \frac{1}{2} v^{2} \Rightarrow v^{2} = (\frac{10}{7}) g h \Rightarrow v = \sqrt{(\frac{10 g h}{7})}$

Page No 200:

Answer:

Let the radius of the disc be R.
Let the angular velocity of the disc ω.
Let the height attained by the disc be h.

On applying the law of conservation of energy, we get:
$\frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2} = m g h \Rightarrow \frac{1}{2} m v^{2} + \frac{1}{2} \times \frac{1}{2} m R^{2} \times {(\frac{v}{R})}^{2} = m g h \Rightarrow \frac{1}{2} v^{2} + \frac{1}{4} v^{2} = g h \Rightarrow \frac{3}{4} v^{2} = g h \Rightarrow h = \frac{3 v^{2}}{4 g}$

Page No 200:

Answer:

Let radius of the sphere be r. Let r be negligible w.r.t. l.

Potential energy of the sphere, P.E. = $m g l \sin θ$
Total kinetic energy of the sphere of mass m rolling with speed v = $\frac{7}{10} m v^{2}$
On applying the law of conservation of energy, we get:
$m g l \sin θ = \frac{7}{10} m v^{2} \Rightarrow g l \sin θ = \frac{7}{10} ν^{2} \Rightarrow ν = \sqrt{(\frac{10}{7} g l \sin θ)}$

Page No 200:

Answer:

It is given that a hollow sphere is released from the top of an inclined plane of inclination θ.
(a) To prevent sliding, the body will make only perfect rolling. In this condition, we have:

$m g l \sin θ - f = m a . . . (1) f \times R = (\frac{2}{3}) m R^{2} \times (\frac{a}{R}) \Rightarrow f = \frac{2}{3} m a . . . (2)$

On putting this value in the equation (1), we get:
$m g \sin θ - \frac{2}{3} m a = m a$
$\Rightarrow a = \frac{3}{5} g \sin θ$
From equation (1), we have:

$m g \sin θ - f = \frac{3}{5} m g \sin θ \Rightarrow f = \frac{2}{5} m g \sin θ \Rightarrow μ m g \cos θ = \frac{2}{5} m g \sin θ \Rightarrow μ = \frac{2}{5} \tan θ$

(b)
$(\frac{1}{5}) \tan θ (m g \cos θ) R = \frac{2}{3} m R^{2} α \Rightarrow α = \frac{3}{10} (\frac{g \sin θ}{R})$
$a_{c} = g \sin θ - (\frac{g}{5}) \sin θ = (\frac{4}{5}) g \sin θ \Rightarrow t^{2} = \frac{2 l}{a_{c}} = 2 l (4 g \frac{\sin θ}{5}) (\frac{5}{2 g \sin θ})$

$∴ ω = a t K . E . = \frac{1}{2} m ν^{2} + \frac{1}{2} I ω^{2} = \frac{1}{2} m (2 a l) + \frac{1}{2} l (a^{2} t^{2}) = \frac{1}{2} m (4 g \frac{\sin θ}{5}) \times 2 \times l + \frac{1}{2} \times \frac{2}{3} m R^{2} \times \frac{9}{100} = (\frac{\sin^{2} θ}{R}) \times (\frac{5 L}{2 g \sin θ}) = 4 m g l \frac{\sin θ}{5} + 3 m g l \frac{\sin θ}{40} = \frac{7}{8} m g l \sin θ$

Page No 200:

Answer:

The radius of the solid sphere (r) is negligible w.r.t. the radius of the hemispherical cup (R).
Let the ball reaches the bottom of the cup with a velocity v and angular velocity ω.
On applying the law of conservation of energy, we get:
$m g R = \frac{1}{2} I ω^{2} + \frac{1}{2} m v^{2}$

$\Rightarrow m g R = \frac{1}{2} \times \frac{2}{5} m ν^{2} + \frac{1}{2} m ν^{2} \Rightarrow \frac{7}{10} m ν^{2} = m g R \Rightarrow ν^{2} = \frac{10}{7} g R$
As shown in the free body diagram, we have:
Total normal force on the ball
$= m g + \frac{m ν^{2}}{R}$
$= m g + \frac{m (\frac{10}{7} g R)}{R} = m g + m g (\frac{10}{7}) = \frac{17}{7} m g$

Page No 200:

Answer:

Let the sphere be thrown with velocity $v'$ and its velocity becomes v at the top-most point.
From the free body diagram of the sphere, at the topmost point, we have:
$\frac{m v^{2}}{R - r} = m g \Rightarrow v^{2} = g (R - r)$

On applying the law of conservation of energy, we have:
$(\frac{1}{2} m ν'^{2} + \frac{1}{2} I ω'^{2}) = 2 m g (R - r) + (\frac{1}{2} m ν^{2} + \frac{1}{2} I ω^{2}) \Rightarrow \frac{7}{10} m ν'^{2} = 2 m g (R - r) + \frac{7}{10} m ν^{2} \Rightarrow \frac{7}{10} m ν'^{2} = 2 m g (R - r) + \frac{7}{10} m g (R - r) \Rightarrow \frac{7}{10} ν'^{2} = \frac{27 g (R - r)}{10} \Rightarrow ν' = \sqrt{\frac{27}{7} g (R - r)}$

Page No 200:

Answer:

(a) Let the velocity and angular velocity of the ball at point A be v and ω, respectively.
Total kinetic energy at point A $= \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}$

Total potential energy at point A $= m g (R + R \sin θ)$
On applying the law of conservation of energy, we have:
Total energy at initial point = Total energy at A
Therefore, we get:
$m g H = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2} + m g R (1 + \sin θ) \Rightarrow m g H - m g R (1 + \sin θ) = \frac{1}{2} m ν^{2} + \frac{1}{2} I ω^{2} \Rightarrow \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2} = m g (H - R - R \sin θ) . . . (1) Total K . E . at A = m g (H - R - R \sin θ)$

(b) Let us now find the acceleration components.
Putting $I = \frac{2}{5} m R^{2} and ω = \frac{v}{R}$ in equation (1), we get:
$\frac{7}{10} m v^{2} = m g (H - R - R \sin θ) \Rightarrow v^{2} = \frac{10}{7} g (H - R - R \sin θ) . . . (2)$
Radial acceleration,
$a_{r} = \frac{v^{2}}{R} = \frac{10}{7} \frac{g (H - R - R \sin θ)}{R}$
For tangential acceleration,
$Differentiating equation (2) w . r . t .' t', 2 v \frac{d v}{d t} = - (\frac{10}{7}) g R \cos θ \frac{d θ}{d t} \Rightarrow ω R \frac{d v}{d t} = - (\frac{5}{7}) g R \cos θ \frac{d θ}{d t} \Rightarrow \frac{d v}{d t} = - (\frac{5}{7}) g c os θ \Rightarrow a_{t} = - (\frac{5}{7}) g c os θ$

(c) At $θ = 0$ , from the free body diagram, we have:

Normal force = $N = m a_{r}$
$N = m \times \frac{10}{7} \frac{g (H - R - R \sin θ)}{R} = (\frac{70}{1000}) \times (\frac{10}{7}) \times 10 \{\frac{0.6 - 0.1}{0.1}\} = 5 N$

At $θ = 0$ , from the free body diagram, we get:
$f_{r} = m g - m a_{t}$ ( $f_{r}$ = Force of friction)
$\Rightarrow f_{r} = m (g - a_{t}) = m (10 - \frac{5}{7} \times 10) = 0.07 (10 - \frac{5}{7} \times 10) = \frac{1}{100} (70 - 50) = 0.2 N$

Page No 200:

Answer:

If the shell does not slip on the surface, its motion should be pure rolling.
Let the cue hits at a height 'h' above the centre.
Let the centre of shell moves with velocity v_c and shell rotates with angular velocity ω after hitting.
For pure rolling, $v_{c} = R ω$
On applying the law of conservation of angular momentum at point O, we get:
$m v_{c} h = I ω m v_{c} h = \frac{2}{3} m R^{2} (\frac{v_{c}}{R}) h = \frac{2 R}{3}$

Page No 200:

Answer:

Initial angular momentum,
$L = I ω = \frac{1}{2} m R^{2} ω$

Angular momentum after it starts pure rolling,
$L' = I ω' + m (v \times R) = \frac{1}{2} m R^{2} (\frac{v}{R}) + m v R = \frac{3}{2} m V R$

As no external torque is applied, angular momentum will be conserved.
Therefore, we have:
L = L'
$\Rightarrow \frac{1}{2} m R^{2} ω = \frac{3}{2} m v R \Rightarrow v = \frac{ω R}{3}$

Page No 200:

Answer:

Initial angular momentum about point A,
$L = m v R$
Angular momentum about point A' after it starts pure rolling,
$L' = I ω + m (v \times R) = \frac{2}{3} m R^{2} (\frac{v'}{R}) + m v' R = \frac{5}{3} m v' R$

As no external torque is applied, angular momentum will be conserved.
Therefore, we have:
L = L'
$\Rightarrow m v R = \frac{5}{3} m v' R \Rightarrow v' = \frac{3 v}{5}$

Page No 200:

Answer:

Let M be the mass of the hollow sphere and α be the angular acceleration produced in the sphere by the tangential force F.

Torque due to this force, $τ = F \times R$
Also, $τ = I α$
$So, F \times R = (\frac{2}{3}) M R^{2} α \Rightarrow α = \frac{3 F}{2 M R}$
Applying $θ = ω_{0} t + \frac{1}{2} α t^{2}$ , we get:
$2 π = \frac{1}{2} α t^{2} \Rightarrow t^{2} = \frac{8 π M R}{3 F}$
Let d be the distance travelled in this time t.
Acceleration, $a = \frac{F}{M}$
$∴ d = \frac{1}{2} a t^{2} = \frac{1}{2} \times \frac{F}{M} \times (\frac{8 π M R}{3 F}) = \frac{4 π R}{3}$

Page No 200:

Answer:

Let α be the angular acceleration produced in the sphere.

Rotational equation of motion,
$F \times R - f_{r} \times R = I α$
$\Rightarrow F = \frac{2}{5} m R α + μ m g . . . (1)$

Translational equation of motion,
$F = m a - μ m g \Rightarrow a = \frac{(F + μ m g)}{m}$
For pure rolling, we have:
$α = \frac{a}{R}$
$\Rightarrow α = \frac{(F + μ m g)}{m R}$

Putting the value of $α$ in equation (1), we get:

$F = \frac{2}{5} \frac{m R (F + μ m g)}{m R} + μ m g \Rightarrow F = \frac{2}{5} (F + μ m g) μ m g \Rightarrow F = \frac{2}{5} F + (\frac{2}{5} \times \frac{2}{7} \times 0.5 \times 10) + (\frac{2}{7} \times 0.5 \times 10) \Rightarrow \frac{3 F}{5} = \frac{4}{7} + \frac{10}{7} = 2 \Rightarrow F = \frac{5 \times 2}{3} = \frac{10}{3} = 3.3 N$

Page No 200:

Answer:

(a)

Initial angular momentum about point A,
$L = m (v \times R) - I ω = m v R - \frac{2}{5} m R^{2} (\frac{v}{R}) = \frac{3}{5} m v R$
Angular momentum about point A' after it stops rotating,
$L' = m v' R$

As no external torque is applied, angular momentum will be conserved.
Therefore, we have:
L = L'
$\Rightarrow m v R = \frac{5}{3} m v' R \Rightarrow v' = \frac{3 v}{5}$

(b)

Angular momentum about point A after it stops rotating,
$L' = m v' R$
Angular momentum about point A' after it starts pure rolling.
$L " = I ω " + m (v " \times R) = \frac{2}{5} m R^{2} (\frac{v "}{R}) + m v " R = \frac{7}{5} m V R$
As no external torque is applied, angular momentum will be conserved.
Therefore, we have:
L' = L"
$\Rightarrow m \frac{3 V}{5} R = \frac{7}{5} m v " R \Rightarrow v " = \frac{3 v}{7}$

Page No 200:

Answer:

Consider two cases:

(a) Just after the collision with the wall, the sphere rebounds with velocity $ν$ towards left but it continues to rotate in the clockwise direction (as shown in figure)

Angular momentum of the sphere about centre,
$L = m v R - I ω = m v R - \frac{2}{5} m R^{2} \times \frac{ν}{R} = \frac{3}{5} m v R$

(b) When pure rolling starts:
Let the velocity be $ν$ ' and the corresponding angular velocity be $\frac{v'}{R}$ .
$L' = m v' R + I ω' = m v' R + \frac{2}{5} m R^{2} \times \frac{v'}{R} = \frac{7}{5} m v' R$
Angular momentum is constant.
Therefore, we have:
$L = L' \Rightarrow v' = \frac{3 v}{7}$
So, the sphere will move with velocity $\frac{3 v}{7}$ .

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