Hc Verma I for Class 11 Science Physics Chapter 7 - Circular Motion

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Hc Verma I Solutions for Class 11 Science Physics Chapter 7 Circular Motion are provided here with simple step-by-step explanations. These solutions for Circular Motion are extremely popular among Class 11 Science students for Physics Circular Motion Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Hc Verma I Book of Class 11 Science Physics Chapter 7 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Hc Verma I Solutions. All Hc Verma I Solutions for class Class 11 Science Physics are prepared by experts and are 100% accurate.

Page No 111:

Answer:

Yes, it is possible to accelerate the motorcycle without putting higher petrol input rate into the engine by driving the motorcycle on a circular track.

Page No 112:

Answer:

The person rotating with the drum will observe that centrifugal force and coriolis force act on the water particles and the person washing the cloth will observe that water particles are thrown outward (away from the drum) and no pseudo force is acting on the particles.

Page No 112:

Answer:

The coin gets the required centripetal force from the frictional force between the coin and the record.

Page No 112:

Answer:

The bird tilts its body and tail in such a way that the air around offers a dragging force in left direction, perpendicular to its initial direction of motion. This dragging force provides the necessary centripetal force to take the left turn.

Page No 112:

Answer:

No, it is not necessary to express all the angles in radian while using the equation ω = ω₀ + at.
If ω (angular velocity) and α (angular acceleration) are in rad/s and rad/s², respectively, we will get the angle in radian.

Page No 112:

Answer:

While shaking, our hand moves on a curved path with some angular velocity and water on our hand feels centrifugal force in the outward direction. Therefore, water get detached from our hand and leaves it.

Page No 112:

Answer:

The outer wall will exert a non-zero normal contact force on the block. As the block moves in a uniform circular motion, centrifugal force in radially outward direction acts on it and it comes in contact with the outer wall of the tube.

Page No 112:

Answer:

(a) Gravitational attraction of the Sun on the Earth is equal to the centripetal force. This statement is more appropriate.

Gravitational attraction of the Sun on the Earth provides the necessary centripetal force required for the circular motion of the Earth around the Sun.

Page No 112:

Answer:

As the wall is at a distance r, the driver should either take a circular turn of radius r or apply brakes to avoid hitting the wall.

Page No 112:

Answer:

When the same mass is set into oscillation, the tension in the string increases because of the additional centripetal force of the mass oscillating in a curved path.

Page No 112:

Answer:

(d) its velocity and acceleration both change.

In a circular motion, the direction of particle changes. Therefore, velocity, being a vector quantity, also changes.
As the velocity changes, acceleration also changes.

Page No 112:

Answer:

(d) 1

Time period (T) is same for both the cars.
We know that:
$ω \propto \frac{1}{T} So, \frac{ω_{1}}{ω_{2}} = 1$

Page No 112:

Answer:

(c) N_A < N_B

From the figure in the question, it is clear that $r_{B} > r_{A}$ .
Here, normal reaction is inversely proportional to the centrifugal force acting on the car, while taking turn on the curve track. Also, centrifugal force is inversely proportional to the radius of the circular track.
Therefore, we have:
N_A < N_B

Page No 112:

Answer:

(d) zero

The centrifugal force is a pseudo force and can only be observed from the frame of reference, which is non-inertial w.r.t. the particle.

Page No 112:

Answer:

(b) $m ω_{0}^{2} a$

The centrifugal force on the particle depends on the angular speed (ω₀) of the frame and not on the angular speed (ω) of the particle. Thus, the value of centrifugal force on the particle is $m ω_{0}^{2} a$ .

Page No 112:

Answer:

(a) 10 cm/s, 10 cm/s²

It is given that the turntable is rotating with uniform angular velocity. Let the velocity be $ω$ .
We have:
$v = r ω \Rightarrow v \propto r (for constant ω) \frac{v}{v'} = \frac{r}{r'} \Rightarrow v' = \frac{v}{2} = 10 cm / s$
Similarly, we have:
$a' = \frac{a}{2} = 10 cm / s^{2}$

Page No 112:

Answer:

(c) mg is not greater than $\frac{m v^{2}}{r}$

At the top of the path, the direction of mg is vertically downward and for centrifugal force $(\frac{m v^{2}}{r})$ , the direction is vertically upward. If the vertically downward force is not greater, water will not fall.

Page No 113:

Answer:

(c) along a tangent

The stone will move in a circle and the direction of velocity at any instant is always along the tangent at that point. Therefore, the stone will move along the tangent to the circle at a point where the string breaks.

Page No 113:

Answer:

(a) 1 cm

Let the force of friction between the coin and the rotating turntable be F.
For the coin to just slip, we have:
$F = m ω^{2} r$
Here, $m ω^{2} r$ is the centrifugal force acting on the coin.
For constant F and m, we have:
$r \propto \frac{1}{ω^{2}}$
Therefore,
$\frac{r'}{r} = {(\frac{ω}{ω'})}^{2} \Rightarrow r' = 1 cm$

Page No 113:

Answer:

(a) increases

The normal force on the motorcycle, $N = m g \cos θ - \frac{m v^{2}}{R}$
As the motorcycle is ascending on the overbridge, θ decreases (from $\frac{π}{2}$ to 0).
So, normal force increases with decrease in θ.

Page No 113:

Answer:

(c) F_C is maximum of the three forces.

At the middle of bridge, normal force can be given as:
$N_{A} = m g$
$N_{B} = \frac{m v^{2}}{r} - m g N_{C} = \frac{m v^{2}}{r} + m g$

So, F_C is maximum.

Page No 113:

Answer:

(a) F₁ > F₂

When the trains are moving, effective angular velocity of both the trains are different (as shown in the figure).

Effective angular velocity of train B is more than that of train A.

Normal force with which both the trains push the tracks is given as:

$N = m g - m R ω'^{2}$

From the above equation, we can conclude that F₁ > F₂.

Page No 113:

Answer:

(d) increase at some places and remain the same at some other places

If the Earth stops rotating on its axis, there will be an increase in the value of acceleration due to gravity at the equator. At the same time, there will be no change in the value of g at the poles.

Page No 113:

Answer:

(a) T₁ > T₂

Let the angular velocity of the rod be $ω$ .
Distance of the centre of mass of portion of the rod on the right side of L/4 from the pivoted end:
$r_{1} = \frac{L}{4} + \frac{1}{2} (\frac{3 L}{4}) = \frac{5 L}{8}$
Mass of the rod on the right side of L/4 from the pivoted end:
$m_{1} = \frac{3}{4} M$
At point L/4, we have:
$T_{1} = m_{1} ω^{2} r_{1} = \frac{3}{4} M ω^{2} \frac{5}{8} L = \frac{15}{32} M ω^{2} L$

Distance of the centre of mass of rod on the right side of 3L/4 from the pivoted end:
$r_{1} = \frac{1}{2} (\frac{L}{4}) + \frac{3 L}{4} = \frac{7 L}{8}$
Mass of the rod on the right side of L/4 from the pivoted end:
$m_{1} = \frac{1}{4} M$
At point 3L/4, we have:
$T_{2} = m_{2} ω^{2} r_{2} = \frac{1}{4} M ω^{2} \frac{7}{8} L = \frac{7}{32} M ω^{2} L$
∴ T₁ > T₂

Page No 113:

Answer:

(d) zero

When the car is in air, the acceleration of bob and car is same. Hence, the tension in the string will be zero.

Page No 113:

Answer:

(c) at the extreme positions

Tension is the string, $T = \frac{m v^{2}}{r} - m g \cos θ$
When v = 0, $|T| = m g \cos θ$
That is, at the extreme positions, the tension is the string is mgcosθ.

Page No 113:

Answer:

(a) speed
(d) magnitude of acceleration

When an object follows a curved path, its direction changes continuously. So, the scalar quantities like speed and magnitude of acceleration may remain constant during the motion.

Page No 113:

Answer:

(d) The instantaneous acceleration of the Earth points towards the Sun.

The speed is constant; therefore, there is no tangential acceleration and the direction of radial acceleration is towards the Sun. So, the instantaneous acceleration of the Earth points towards the Sun.

Page No 113:

Answer:

(b) speed remains constant
(d) tangential acceleration remains constant

If the speed is constant, the position vector of the particle sweeps out equal area in equal time in circular motion.
Also, for constant speed, tangential acceleration is zero, i.e., constant.

Page No 113:

Answer:

(c) The magnitude of acceleration is constant.

As the pitch and radius of the path is constant, it shows that the magnitude of tangential and radial acceleration is also constant.
Hence, the magnitude of total acceleration is constant.

Page No 114:

Answer:

(b) The magnitude of the frictional force on the car is greater than $\frac{m v^{2}}{r}$ .
(c) The friction coefficient between the ground and the car is not less than a/g.

If the magnitude of the frictional force on the car is not greater than $\frac{m v^{2}}{r}$ , it will not move forward, as its speed (v) is increasing at a rate a.

Page No 114:

Answer:

(b) If the car turns at a speed less than 40 km/hr, it will slip down.
(d) If the car turns at the correct speed of 40 km/hr, the force by the road on the car is greater than mg as well as greater than $\frac{m v^{2}}{r}$ .

The friction is zero and the road is banked for a speed v = 40 km/hr. If the car turns at a speed less than 40 km/hr, it will slip down.

Page No 114:

Answer:

(b) There are other forces on the particle.
(d) The resultant of the other forces varies in magnitude as well as in direction.

We cannot move a particle in a circle by just applying a constant force. So, there are other forces on the particle.
As a constant force cannot move a particle in a circle, the resultant of other forces varies in magnitude as well as in direction (because $\vec{F}$ is constant).

Page No 114:

Answer:

Distance between the Earth and the Moon:
$r = 3.85 \times 10^{5} km = 3.85 \times 10^{8} m$

Time taken by the Moon to revolve around the Earth:

$T = 27.3 days = 24 \times 3600 \times 27.3 s = 2.36 \times 10^{6} s Velocity of the Moon : v = \frac{2 π r}{T} = \frac{2 \times 3.14 \times 3.85 \times 10^{8}}{2.36 \times 10^{6}} = 1025.42 m / s Acceleration of the Moon : a = \frac{v^{2}}{r} = \frac{(1025.42)^{2}}{2.36 \times 10^{6}} = 0.00273 m / s^{2}$
$\Rightarrow a = 2.73 \times 10^{- 3} m / s^{2}$

Page No 114:

Answer:

Diameter of the Earth = 12800 km
So, radius of the Earth, R = 6400 km = 6.4 × 10⁶ m

Time period of revolution of the Earth about its axis:

$T = 24 hr = 24 \times 3600 s v = \frac{2 π r}{T} = \frac{2 \times 3.14 \times 64 \times 10^{6}}{24 \times 3600}$ ^{$\Rightarrow v = 465.185 m / s$}
$Acceleration of the particle : a = \frac{v^{2}}{R} = \frac{{(465.185)}^{2}}{64 \times 10^{5}} = 0.038 m / s^{2}$

Page No 114:

Answer:

Speed is given as a function of time. Therefore, we have:
v = 2t
Radius of the circle = r = 1 cm
At time t = 2 s, we get:
(a) Radial acceleration
$a = \frac{v^{2}}{r} = \frac{2^{2}}{1} = 4 cm / s^{2}$
(b) Tangential acceleration
$a = \frac{d v}{d t} = \frac{d}{d t} (2 t) = 2 cm / s^{2}$
(c) Magnitude of acceleration
$a = \sqrt{4^{2} + 2^{2}} = \sqrt{20} cm / s^{2}$

Page No 114:

Answer:

Given:
Mass = m = 150 kg
Speed = v = 36 km/hr = 10 m/s
Radius of turn = r = 30 m
Let the horizontal force needed to make the turn be F. We have:
$F = \frac{m v^{2}}{r} = \frac{150 \times (10)^{2}}{30} = \frac{150 \times 100}{30} = 500 N$

Page No 114:

Answer:

Given:
Speed of the scooter = v = 36 km/hr = 10 m/s
Radius of turn = r = 30 m
Let the angle of banking be $θ$ . We have:
$\tan θ = \frac{v^{2}}{r g}$
$\Rightarrow \tan θ = \frac{100}{30 \times 10} = \frac{1}{3} \Rightarrow θ = \tan^{- 1} (\frac{1}{3})$

Page No 114:

Answer:

Given:
Speed of the vehicle = v = 18 km/h = 5 m/s
Radius of the park = r = 10 m
Let the angle of banking be $θ$ .
Thus, we have:
$\tan θ = \frac{v^{2}}{r g}$
$\Rightarrow θ = \tan^{- 1} (\frac{25}{100}) \Rightarrow θ = \tan^{- 1} (\frac{1}{4})$

Page No 114:

Answer:

If the road is horizontal (no banking), we have:
$\frac{m v^{2}}{R} = f_{s} N = m g$
Here, f_s is the force of friction and N is the normal reaction.
If μ is the friction coefficient, we have:
$Friction force = f_{s} = μ N So, \frac{m v^{2}}{R} = μ m g$

Here,
Velocity = v = 5 m/s
Radius = R = 10 m
$∴ \frac{25}{10} = μ g \Rightarrow μ = (\frac{25}{100}) = 0.25$

Page No 114:

Answer:

Given:
Angle of banking = θ = 30°
Radius = r = 50 m
Assume that the vehicle travels on this road at speed v so that the friction is not used.
We get:
$\tan θ = \frac{v^{2}}{r g} \Rightarrow \tan 30 ° = \frac{v^{2}}{r g} \Rightarrow \frac{1}{\sqrt{3}} = \frac{v^{2}}{r g} \Rightarrow v^{2} = \frac{r g}{\sqrt{3}} = \frac{50 \times 10}{\sqrt{3}} \Rightarrow v = \sqrt{\frac{500}{\sqrt{3}}} = 17 m / s$

Page No 114:

Answer:

$Given : Radius of the orbit of the ground state = r = 5.3 \times 10^{- 11} m Mass of the electron = m = 9.1 \times 10^{- 31} kg Charge of electron = q = 1.6 \times 10^{- 19} c We know : Centripetal force = Coulomb attraction Therefore, we have : \Rightarrow \frac{m v^{2}}{r} = \frac{1}{4 {πε}_{\circ}} \frac{q^{2}}{r^{2}} \Rightarrow v^{2} = \frac{1}{4 {πε}_{\circ}} \frac{q^{2}}{r m} = \frac{9 \times 10^{9} \times {(1.6 \times 10^{- 19})}^{2}}{5.3 \times 10^{- 11} \times 9.1 \times 10^{- 31}} = \frac{23.04}{48.23} \times 10^{13} = 0.477 \times 10^{13} = 4.7 \times 10^{12} \Rightarrow v = \sqrt{4.7} \times 10^{12} = 2.2 \times 10^{6} m / s$

Page No 114:

Answer:

Let m be the mass of the stone.
Let v be the velocity of the stone at the highest point.
R is the radius of the circle.
Thus, in a vertical circle and at the highest point, we have:
$\frac{m v^{2}}{R} = m g \Rightarrow v^{2} = R g \Rightarrow v = \sqrt{R g}$

Page No 114:

Answer:

Diameter of the fan = 120 cm
∴ Radius of the fan = r = 60 cm = 0.6 m
Mass of the particle = M = 1 g = 0.001 kg
Frequency of revolutions = n = 1500 rev/min = 25 rev/s
Angular velocity = ω = 2 $π$ n = 2 $π$ × 25 = 157.14 rev/s
Force of the blade on the particle:
F = Mrw²
= (0.001) × 0.6 × (157.14)²
=14.8 N
The moving fan exerts this force on the particle.
The particle also exerts a force of 14.8 N on the blade along its surface.

Page No 114:

Answer:

$Frequency of disc = n = 33 \frac{1}{3} rev / m = \frac{100}{3 \times 60} rev / s$
$Angular velocity = ω = 2 π n = 2 π \times \frac{100}{180} = \frac{10 π}{9} rad / s$
$r = 10 cm = 0.1 m g = 10 m / s^{2}$

It is given that the mosquito is sitting on the L.P. record disc. Therefore, we have:
Friction force ≥ Centrifugal force on the mosquito
⇒ μmg ≥ mrω²
⇒ μ ≥ rω²/g
$\Rightarrow μ \geq 0.1× {(\frac{10π}{9})}^{2} \frac{1}{10} \Rightarrow μ \geq \frac{π^{2}}{81}$

Page No 114:

Answer:

Speed of the car = v = 36 km/hr = 10 m/s
Acceleration due to gravity = g = 10 m/s²

Let T be the tension in the string when the pendulum makes an angle θ with the vertical.
From the figure, we get:
$T \sin θ = \frac{m v^{2}}{r} . . . (i) T \cos θ = m g . . . (ii) \Rightarrow \frac{\sin θ}{\cos θ} = \frac{m v^{2}}{r m g} \Rightarrow \tan θ = \frac{v^{2}}{r g} \Rightarrow θ = \tan^{- 1} (\frac{v^{2}}{r g}) = \tan^{- 1} [\frac{100}{(10 \times 10)}] = \tan^{- 1} (1) \Rightarrow θ = 45 °$

Page No 115:

Answer:

Given:
Mass of the bob = m = 100 gm = 0.1 kg
Length of the string = r = 1 m
Speed of bob at the lowest point in its path = 1.4 m/s
Let T be the tension in the string.
From the free body diagram, we get:
$T = m g + \frac{m v^{2}}{r} = (\frac{1}{10}) \times 9.8 + \frac{(1.4)^{2}}{10} = 0.98 + 0.196 = 1.176 \approx 1.2 N$

Page No 115:

Answer:

Given:
Mass of the bob = m = 0.1 kg
Length of the circle = R = 1 m
Velocity of the bob = v = 1.4 m/s
Let T be the tension in the string when it makes an angle of 0.20 radian with the vertical.

From the free body diagram, we get:
$T - m g \cos θ = \frac{m v^{2}}{R} T = \frac{m v^{2}}{R} + m g \cos θ For small θ, it is given that : \cos θ = 1 - \frac{θ^{2}}{2} ∴ T = \frac{0.1 \times (1.4)^{2}}{1} + (0.1) \times 9.8 (1 - \frac{θ^{2}}{2}) = 0.196 + 0.98 \times (1 - \frac{{(0.2)}^{2}}{2}) = 0.196 + 0.9604 = 1.156 N \approx 1.16 N$

Page No 115:

Answer:

Let T be the tension in the string at the extreme position.
Velocity of the pendulum is zero at the extreme position.
So, there is no centripetal force on the bob.
∴ T = mgcosθ₀

Page No 115:

Answer:

(a) Balance reading = Normal force on the balance by the Earth.
At equator, the normal force (N) on the spring balance:
N = mg − mω²r

True weight = mg
Therefore, we have:
$Fraction less than the true weight = \frac{m g - (m g - m ω^{2} r)}{m g} = \frac{ω^{2} r}{g} = {(\frac{2 π}{24 \times 3600})}^{2} (\frac{6.4 \times 10^{6}}{10})$
$= 3.5 \times 10^{- 3}$

(b) When the balance reading is half, we have:
$True weight = \frac{m g - m ω^{2} r}{m g} = \frac{1}{2}$

$\Rightarrow ω^{2} r = \frac{g}{2} \Rightarrow ω = \sqrt{\frac{g}{2 r}} = \sqrt{\frac{10}{2 \times 6400 \times 10^{3}}} rad / s$

$∴ Duration of the day = 2 π \times \sqrt{\frac{2 \times 6400 \times 10^{3}}{9.8}} s = 2 π \sqrt{\frac{6.4 \times 10^{7}}{49}} s = \frac{2 π \times 8000}{7 \times 3600} h = 2 h$

Page No 115:

Answer:

Given:
Speed of vehicles = v = 36 km/hr = 10 m/s
Radius = r = 20 m
Coefficient of static friction = μ = 0.4
Let the road be banked with an angle $θ$ . We have:
$θ = \tan^{- 1} \frac{v^{2}}{r g} = \tan^{- 1} \frac{100}{20 \times 10} = \tan^{- 1} (\frac{1}{2}) \Rightarrow tanθ = 0.5$

When the car travels at the maximum speed, it slips upward and μN₁ acts downward.
Therefore we have:
$N_{1} - m g \cos θ - \frac{m v_{1}^{2}}{r} \sin θ = 0 . . . (i) μ N_{1} + m g \sin θ - \frac{m v_{1}^{2}}{r} \cos θ = 0 . . . (ii)$

On solving the above equations, we get:
$v_{1} = \sqrt{r g \frac{μ + \tan θ}{1 - μ \tan θ}} = \sqrt{20 \times 10 \times \frac{0.9}{0.8}} = 15 m / s = 54 km / hr$

Similarly, for the other case, it can be proved that:
$v_{2} = \sqrt{r g \frac{tanθ - μ}{\sqrt{1 - μ tanθ}}} = \sqrt{20 \times 10 \times \frac{0.1}{1.2}} = 4.08 m / s = 14.7 km / hr$

Thus, the possible speeds are between 14.7 km/hr and 54 km/hr so that the car neither slips down nor skids up.

Page No 115:

Answer:

R = Radius of the bridge
L = Total length of the over bridge

(a) At the highest point:
Let m be the mass of the motorcycle and v be the required velocity.

$m g = \frac{m v^{2}}{R} \Rightarrow v^{2} = R g \Rightarrow v = \sqrt{R r g}$

$(b) Given : v = (\frac{1}{\sqrt{2}}) \sqrt{R g}$

Suppose it loses contact at B.
$At point B, we get : m g \cos θ = \frac{m v^{2}}{R} \Rightarrow v^{2} = R g \cos θ Putting the value of v, we get : \sqrt{{(\frac{R g}{2})}^{2}} = R g \cos θ \Rightarrow \frac{R g}{2} = R g \cos θ \Rightarrow \cos θ = \frac{1}{2} \Rightarrow θ = 60 ° = \frac{π}{3} ∵ θ = \frac{L}{R} ∴ L = R θ = \frac{π R}{3}$
So, it will lose contact at a distance $\frac{π R}{3}$ from the highest point.
(c) Let the uniform speed on the bridge be v. The chances of losing contact is maximum at the end bridge. We have:

$α = \frac{L}{2 R} So, \frac{m v^{2}}{R} = m g \cos α \Rightarrow v = \sqrt{g Rcos (\frac{L}{2 R})}$

Page No 115:

Answer:

Let v be the speed of the car.
Since the motion is non-uniform, the acceleration has both radial (a_r) and tangential (a_t) components.
$a_{r} = \frac{v^{2}}{R} a_{t} = \frac{d v}{d t} = a Resultant magnitude = \sqrt{{(\frac{v^{2}}{R})}^{2} + a^{2}}$

From free body diagram, we have:
$m N = m \sqrt{{(\frac{v^{2}}{R})}^{2} + a^{2}} \Rightarrow μ m g = m \sqrt{{(\frac{v^{2}}{R})}^{2} + a^{2}} \Rightarrow μ^{2} g^{2} = \frac{v^{4}}{R^{2}} + a^{2}$
$\Rightarrow \frac{v^{4}}{R^{2}} = (μ^{2} g^{2} - a^{2}) \Rightarrow v^{4} = (μ^{2} g^{2} - a^{2}) R^{2} \Rightarrow v = [(μ^{2} g^{2} - a^{2}) R^{2}]^{1 / 4}$

Page No 115:

Answer:

(a) Given:
Mass of the block = m
Friction coefficient between the ruler and the block = μ
Let the maximum angular speed be ω₁for which the block does not slip
Now, for the uniform circular motion in the horizontal plane, we have:
$μ m g = m {ω_{1}}^{2} L ∴ ω_{1} = \sqrt{\frac{μ g}{L}}$

(b) Let the block slip at an angular speed ω₂_.
For the uniformly accelerated circular motion, we have:
$μ m g = \sqrt{{(m ω_{2}^{2} L)}^{2} + {(m L α^{2})}^{2}} \Rightarrow ω_{2}^{4} + α^{2} = \frac{μ^{2} g^{2}}{L^{2}}$
$\Rightarrow ω_{2} = {[{(\frac{μ g}{L})}^{2} - α^{2}]}^{1 / 4}$

Page No 115:

Answer:

Given:
Radius of the curves = r = 100 m
Mass of the cycle = m = 100 kg
Velocity = v = 18 km/hr = 5 m/s

$(a) At B, we have : m g - \frac{m v^{2}}{r} = N \Rightarrow N = (100 \times 10) - (100 \times \frac{25}{100}) = 1000 - 25 = 975 N At D, we have : N = m g + \frac{m v^{2}}{r} = 1000 + 25 = 1025 N$

(b) At B and D, we have:
Tendency of the cycle to slide is zero.
So, at B and D, frictional force is zero.
At C, we have:
mgsinθ = f
$\Rightarrow 1000 \times (\frac{1}{\sqrt{2}}) = 707 N$
$(c) (i) Before C, m g \cos θ - N = \frac{m v^{2}}{r} \Rightarrow N = m g \cos θ - \frac{m v^{2}}{r} = 707 - 25 = 682 N (i i) N - m g \cos θ = \frac{m v^{2}}{r} \Rightarrow N = \frac{m v^{2}}{r} + m g \cos θ = 25 + 707 = 732 N$

(d) To find the minimum coefficient of friction, we have to consider a point where N is minimum or a point just before c .
$Therefore, we have : μ N = m g \sin θ \Rightarrow μ \times 682 = 707 \Rightarrow μ = 1.037$

Page No 115:

Answer:

Given:
$Frequency of rod = n = 20 rev per \min \Rightarrow n = \frac{20}{60} = \frac{1}{3} rev / s$
Therefore, we have:
angular velocity of rod,
$Angular velocity of rod = ω = 2 π n = \frac{2 π}{3} rad / s$
Mass of each kid = $m = 15 kg$
Radius = $r = \frac{3}{2} = 1.5 m$

$∴ Frictional force = F = m r ω^{2} \Rightarrow F = 15 \times (1.5) \times \frac{(2 π)^{2}}{9} = 5 \times (0.5) \times 4 π^{2} = 10 π^{2} N$
Thus, the force of frictional on one of the kids is 10 $π$ ².

Page No 115:

Answer:

When the bowl rotates at maximum angular speed, the block tends to slip upwards.
Also, the frictional force acts downward.
Here, we have:
Radius of the path that the block follow = r = Rsinθ
Let N₁ be the normal reaction on the block and ω₁ be the angular velocity after which the block will slip.
From the free body diagram-1, we get:
$N_{1} - m g \cos θ = {mω}_{1}^{2} (R \sin θ) \sin θ . . . (i) μ N_{1} + m g sinθ = m ω_{1}^{2} (R \sin θ) \cos θ . . . (ii)$

On solving the two equation, we get:
$ω_{1} = {[\frac{g (sinθ + μ cosθ)}{R sinθ (cosθ - μ sinθ)}]}^{1 / 2}$

Let us now find the minimum speed (ω₂) on altering the direction of $μ$ (as shown in figure):
$ω_{2} = {[\frac{g (sinθ - μ cosθ)}{R sinθ (cosθ + μ sinθ)}]}^{1 / 2}$

Hence, the range of speed is between ω₂ and ω₁.

Page No 115:

Answer:

At the highest point, the vertical component of velocity is zero.
So, at the highest point, we have:
velocity = v = ucosθ
Centripetal force on the particle = $\frac{m v^{2}}{r}$
$\Rightarrow \frac{m v^{2}}{r} = \frac{m u^{2} \cos^{2} θ}{r}$
At the highest point, we have:
$m g = \frac{m v^{2}}{r}$
Here, r is the radius of curvature of the curve at the point.
$\Rightarrow r = \frac{u^{2} \cos^{2} θ}{g}$

Page No 115:

Answer:

Let u be the initial velocity and v be the velocity at the point where it makes an angle $\frac{θ}{2}$ with the horizontal component.
It is given that the horizontal component remains unchanged.
Therefore, we get:
$v \cos (\frac{θ}{2}) = u cosθ$

$\Rightarrow v = \frac{u \cos θ}{\cos \frac{θ}{2}} . . . (i) m g \cos \frac{θ}{2} = \frac{m v^{2}}{r} . . . (ii) \Rightarrow r = \frac{v^{2}}{g \cos \frac{θ}{2}}$
On substituting the value of v from equation (i), we get:
$r = \frac{u^{2} \cos^{2} θ}{g \cos^{2} \frac{θ}{2}}$

Page No 115:

Answer:

Given:
Radius of the room = R
Mass of the block = m
(a) Normal reaction by the wall on the block = N = $\frac{m v^{2}}{R}$
(b) Force of frictional by the wall = $μ N = \frac{μ m v^{2}}{R}$
(c) Let a_t be the tangential acceleration of the block.
From figure, we get:
$- \frac{μ m v^{2}}{R} = m a_{t} \Rightarrow a_{t} = - \frac{μ v^{2}}{R}$
(d)
$On using a = \frac{d v}{d t} = v \frac{d v}{d s}, we get : v \frac{d v}{d s} = \frac{μ v^{2}}{R} \Rightarrow d s = - \frac{R}{μ} \frac{d v}{v} Integrating both side, we get : s = - \frac{R}{μ} In v + c At, s = 0, v = v_{0} S o, c = \frac{R}{μ} In v_{0} \Rightarrow s = - \frac{R}{μ} In \frac{v}{v_{0}} \Rightarrow \frac{v}{v_{0}} = e^{- \frac{μ s}{R}} \Rightarrow v = v_{0} e^{- \frac{μs}{R}} For one rotation, we have : s = 2 π r ∴ v = v_{0} e^{- 2 π μ}$

Page No 116:

Answer:

Let the mass of the particle be m.
Radius of the path = R
Angular velocity = ω
Force experienced by the particle = mω²R
The component of force mRω² along the line AB (making an angle with the radius) provides the necessary force to the particle to move along AB.
$∴ m ω^{2} R \cos θ = m a \Rightarrow a = ω^{2} R \cos θ$
Let the time taken by the particle to reach the point B be t.
$On using equation of motion, we get : L = u t + \frac{1}{2} a t^{2} \Rightarrow L = \frac{1}{2} ω^{2} R \cos θ t^{2} \Rightarrow t^{2} = \frac{2 L}{ω^{2} R \cos θ} \Rightarrow t = \sqrt{\frac{2 L}{ω^{2} R \cos θ}}$

Page No 116:

Answer:

Given:
Speed of the car = v = 36 km/h = 10 m/s
Radius of the road = r = 50 m
Friction coefficient between the block and the plate = μ = 0.58
Mass of the small body = m = 100 g = 0.1 kg

(a) Let us find the normal contact force (N) exerted by the plant of the block.
$N = \frac{m v^{2}}{r} = 0.1 \times \frac{100}{50} = \frac{1}{5} = 0.2$

(b) The plate is turned; so, the angle between the normal to the plate and the radius of the rod slowly increases.
Therefore, we have:
$N = \frac{m v^{2}}{r} cosθ . . . (i) μ N = \frac{m v^{2}}{r} sinθ . . . (i i) On using (i) and (ii), we get : \frac{μ m v^{2}}{r} \cos θ = \frac{m v^{2}}{r} \sin θ \Rightarrow μ = tanθ \Rightarrow θ = \tan^{- 1} (0.58) \approx 30 °$

Page No 116:

Answer:

Let the bigger mass accelerates towards left with acceleration a.
Let T be the tension in the string and ω be the angular velocity of the table.
From the free body diagram, we have:
$T - m a - m ω^{2} R = 0 . . . (i) T + 2 m a - 2 m ω^{2} R = 0 . . . (i i) On subtracting e q . (i) by e q . (i i), we get : 3 m a = m ω^{2} R \Rightarrow a = \frac{ω^{2} R}{3}$

Substituting the value of a in eq. (i), we get:
$T = m (\frac{ω^{2} R}{3}) + m ω^{2} R \Rightarrow T = \frac{4}{3} m ω^{2} R$

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