# HC Verma Class XII Physics Chapter 7– Circular Motion

#### Exercise : Solution of Questions on page Number : 112

#### Answer: 2

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.

#### Answer: 3

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

#### Answer: 4

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.

#### Answer: 5

No, it is not necessary to express all the angles in radian while using the equation *ω = **ω*_{0} + *at*.

If *ω *(angular velocity) and *α* (angular acceleration) are in rad/s and rad/s^{2}, respectively, we will get the angle in radian.

#### Answer: 6

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.

#### Answer: 7

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.

#### Answer: 8

(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.

#### Answer: 9

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.

#### Answer: 10

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.

#### Answer: 1

(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.

#### Answer: 2

(d) 1

Time period (*T*) is same for both the cars.

We know that : ω∝1T

So, ω1ω2 = 1

#### Answer: 3

(c) *N _{A}* < N

_{B}

From the figure in the question, it is clear that rB > rA.

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 }

#### Answer: 4

(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.

#### Answer: 5

(b) mω02a

The centrifugal force on the particle depends on the angular speed (*ω*_{0}) of the frame and not on the angular speed (*ω*) of the particle. Thus, the value of centrifugal force on the particle is mω02a.

#### Answer: 6

(a) 10 cm/s, 10 cm/s^{2}

It is given that the turntable is rotating with uniform angular velocity. Let the velocity be *ω*.

We have:

v = rω

⇒ v∝r (for constant ω)vv’= rr’

⇒ v’= v2 = 10 cm/s

Similarly, we have:

a’= a^{2 }= 10 cm/s^{2}

#### Answer: 7

(c) *mg* is not greater than mv2r

At the top of the path, the direction of *mg *is vertically downward and for centrifugal force mv2r, the direction is vertically upward. If the vertically downward force is not greater, water will not fall.

#### Exercise : Solution of Questions on page Number : 113

#### Answer: 8

(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.

#### Answer: 9

(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 = mω2r

Here, mω2r is the centrifugal force acting on the coin.

For constant *F* and *m*, we have:

r∝1ω2

Therefore,

r’r = ωω’2

⇒ r’= 1 cm

#### Answer: 10

(a) increases

The normal force on the motorcycle, N = mgcosθ – mv2R

As the motorcycle is ascending on the overbridge, *θ* decreases (from π2 to 0).

So, normal force increases with decrease in *θ*.

#### Answer: 11

(c) *F*_{C} is maximum of the three forces.

At the middle of bridge, normal force can be given as:

NA = mg

NB = mv2r – mgNC = mv2r + mg

So, *F*_{C} is maximum.

#### Answer: 12

(a) *F*_{1} > *F*_{2}

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 = mg-mRω’2

From the above equation, we can conclude that *F*_{1} > *F*_{2}.

#### Answer: 13

(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.

#### Answer: 14

(a) *T*_{1} > *T*_{2}

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:

r1 = L4 + 123L4 = 5L8

Mass of the rod on the right side of *L*/4 from the pivoted end:

m1 = 34M

At point *L*/4, we have:

T1 = m1ω2r1 =34 Mω258L = 1532 Mω2L

Distance of the centre of mass of rod on the right side of 3*L*/4 from the pivoted end:

r1 = 12L4 + 3L4 = 7L8

Mass of the rod on the right side of *L*/4 from the pivoted end:

m1 = 14M

At point 3*L*/4.

we have:

T2 = m2ω2r2 = 14Mω278L = 732Mω2L

∴ *T*_{1} > *T*_{2 }

#### Answer: 15

(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.

#### Answer: 16

(c) at the extreme positions

Tension is the string, T = mv2r – mgcosθ

When *v = *0, T = mgcosθ

That is, at the extreme positions, the tension is the string is *mg*cos*θ**. *

#### Answer: 1

(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.

#### Answer: 2

(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.

#### Answer: 3

(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.

#### Answer: 4

(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.

#### Exercise : Solution of Questions on page Number : 114

#### Answer: 5

(b) The magnitude of the frictional force on the car is greater than mv2r.

(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 mv2r, it will not move forward, as its speed (*v*) is increasing at a rate *a*.

#### Answer: 6

(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 mv2r.

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.

#### Answer: 7

(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 F→ is constant).

#### Answer: 1

Distance between the Earth and the Moon:

r = 3.85 × 105 km = 3.85 × 108 m

Time taken by the Moon to revolve around the Earth:

T = 27.3 days = 24×3600 × 27.3 s = 2.36 × 106 s

Velocity of the Moon : v = 2πrT = 2 × 3.14 × 3.85 × 1082.36 × 106 = 1025.42 m/s

Acceleration of the Moon : a = v2r = (1025.42)22.36 × 106 = 0.00273 m/s2

⇒ a = 2.73 × 10 – 3 m/s2

#### Answer: 2

Diameter of the Earth = 12800 km

So, radius of the Earth, R = 6400 km = 6.4 × 10^{6} m

Time period of revolution of the Earth about its axis:

T = 24 hr = 24 × 3600 sv = 2πrT = 2 × 3.14 × 64 × 10624 × 3600 ^{ ⇒v = 465.185 m/s}

Acceleration of the particle :

a = v2R = 465.18526 4× 105 = 0.038 m/s2

#### Answer: 3

Speed is given as a function of time.

Therefore, we have : *v* = 2*t*

Radius of the circle = *r* = 1 cm

At time *t *= 2 s, we get:

(a) Radial acceleration

a = v2r = 221 = 4 cm/s2

(b) Tangential acceleration

a = dv,dt = ddt2t = 2 cm/s2

(c) Magnitude of acceleration

a = 42 + 22 = 20 cm/s2

#### Answer: 4

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 = mv2r = 150 × (10)230 = 150 × 10030 = 500 N

#### Answer: 5

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 : tanθ = v2rg

⇒ tanθ = 10030 × 10 = 13

⇒ θ = tan-113

#### Answer: 6

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:

tanθ = v2rg

⇒ θ = tan-125100

⇒ θ = tan-114

#### Answer: 7

If the road is horizontal (no banking), we have:

mv2R = fsN = mg

Here, *f _{s}* is the force of friction and

*N*is the normal reaction.

If

*μ*is the friction coefficient, we have:

Friction force = fs = μNSo,

mv2R = μmg

Here,

Velocity =

*v*= 5 m/s

Radius

*= R*= 10 m

∴ 2510 = μg

⇒ μ = 25100 = 0.25

#### Answer: 8

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 : tanθ = v2rg

⇒ tan 30°= v2rg

⇒ 13 = v2rg

⇒ v2 = rg3 = 50 × 103

⇒ v = 5003 = 17 m/s

#### Answer: 9

Given:Radius of the orbit of the ground state = r = 5.3 × 10-11 m

Mass of the electron = m = 9.1 × 10-31 kg

Charge of electron = q = 1.6×1019 c

We know:Centripetal force=Coulomb attraction Therefore,

we have :

⇒ mv2r = 14πε∘q2r2

⇒ v2 = 14πε∘q2rm = 9 × 109 × 1.6 × 10 – 1925.3 × 10 – 11 × 9.1 × 10 – 31 = 23.0448.23 × 1013 = 0.477 × 1013 = 4.7 × 1012

⇒ v = 4.7 × 1012 = 2.2 × 106 m/s

#### Answer: 10

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:

mv2R = mg

⇒ v2 = Rg

⇒ v = Rg

#### Answer: 11

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*πn* = 2π × 25 = 157.14 rev/s

Force of the blade on the particle:

*F* = M*rw*^{2}

= (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.

#### Answer: 12

Frequency of disc = n = 3313rev/m = 1003 × 60rev/s

Angular velocity = ω = 2πn = 2π × 100180 = 10π9rad/s

r = 10 cm = 0.1 mg = 10 m/s2

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* ≥ *mrω*^{2}

*⇒ μ *≥ *rω*^{2}*/g*

⇒ μ ≥ 0.1×10π92110

⇒ μ ≥ π281

#### Answer: 13

Speed of the car = *v* = 36 km/hr =10 m/s

Acceleration due to gravity* = g*= 10 m/s^{2}

Let *T* be the tension in the string when the pendulum makes an angle* θ* with the vertical.

From the figure, we get:

Tsinθ = mv2r …(i)

Tcosθ = mg …(ii)

⇒ sinθcosθ = mv2rmg

⇒ tanθ = v2rg

⇒ θ = tan – 1v2rg = tan-1100(10×10) = tan-1(1)

⇒ θ = 45°

#### Exercise : Solution of Questions on page Number : 115

#### Answer: 14

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 = mg + mv2r = 110×9.8 + (1.4)210 = 0.98+0.196 = 1.176 ≈ 1.2 N

#### Answer: 15

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-mgcosθ=mv2RT=mv2R+mg cosθFor small θ, it is given that :

cosθ = 1-θ22

∴ T = 0.1 × (1.4)21 + (0.1) × 9.81 – θ22 = 0.196 + 0.98 × 1-0.222 = 0.196 + 0.9604 = 1.156 N≈1.16 N

#### Answer: 16

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* = *mg*cos*θ*_{0}

#### Answer: 17

(a) Balance reading = Normal force on the balance by the Earth.

At equator, the normal force (N) on the spring balance:

N *=* *mg* − *mω*^{2}*r*

True weight =* mg*

Therefore, we have:

Fraction less than the true weight

= mg-(mg-mω2r)mg

= ω2rg

= 2π24 × 360026.4 × 10610

= 3.5 × 10-3

(b) When the balance reading is half, we have:

True weight = mg – mω2rmg = 12

⇒ ω2r = g2

⇒ ω = g2r = 102×6400×103 rad/s

∴ Duration of the day = 2π × 2 × 6400 × 1039.8 s = 2π6.4 × 10749s = 2π × 80007 × 3600 h = 2 h

#### Answer: 18

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-1v2rg

= tan – 110020×10

= tan – 112

⇒ tanθ = 0.5

When the car travels at the maximum speed, it slips upward and *μN*_{1} acts downward.

Therefore we have:

N1 – mgcosθ – mv12rsinθ = 0 …(i)

μN1 + mgsinθ – mv12rcosθ = 0 …(ii)

On solving the above equations, we get:

v1 = rgμ + tanθ1 – μtanθ

= 20 × 10 × 0.90.8 = 15 m/s

= 54 km/hr

Similarly, for the other case, it can be proved that:

v2 = rgtanθ – μ1 – μ tanθ = 20 × 10 × 0.11.2 = 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.

#### Answer: 19

*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.

#### Answer: 20

Let *v* be the speed of the car.

Since the motion is non-uniform, the acceleration has both radial (*a _{r}*) and tangential (

*a*) components.

_{t}ar = v2Rat = dv dt = a

Resultant magnitude = v2R2 + a2

From free body diagram, we have:

mN = mv2R2 + a2

⇒ μmg = mv2R2 + a2

⇒ μ2g2 = v4R2 + a2

⇒ v4R2= (μ2g2 – a2)

⇒ v4 = (μ2g2-a2)R2

⇒ v = [(μ2g2-a2)R2] 1/4

#### Answer: 21

(a) Given:

Mass of the block = *m*

Friction coefficient between the ruler and the block = *μ*

Let the maximum angular speed be *ω*_{1}_{ }for which the block does not slip

Now, for the uniform circular motion in the horizontal plane, we have:

μmg = mω12L

∴ω1 = μgL

(b) Let the block slip at an angular speed *ω*_{2}_{.}_{ }

For the uniformly accelerated circular motion, we have:

μmg = mω22L2 + mLα22

⇒ ω24 + α2 = μ2g2L2

⇒ ω2 = μgL2 – α21/4

#### Answer: 22

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 :

mg – mv2r = N

⇒ N = (100×10) – 100×25100 = 1000-2 5 = 975 NAtD,

we have: N = mg + mv2r = 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:

*mg*sin*θ* = *f*

⇒ 1000 × 12 = 707 N

(c) (i) Before C, mgcosθ-N=mv2r

⇒ N = mg cosθ – mv2r = 707 – 25 = 682 N

(ii) N-mgcos θ= mv2r

⇒ N = mv2r + mgcosθ = 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:

μN = mgsinθ

⇒ μ × 682 = 707

⇒ μ = 1.037

#### Answer: 23

Given:

Frequency of rod=n=20 rev per min

⇒ n= 2060 = 13rev/s

Therefore, we have : angular velocity of rod,

Angular velocity of rod = ω = 2πn = 2π3rad/s

Mass of each kid = m = 15 kg

Radius = r = 32 = 1.5 m

∴Frictional force = F = mrω2

⇒ F = 15 × (1.5) × (2π)29 = 5 × (0.5) × 4π2 = 10π2 N

Thus, the force of frictional on one of the kids is 10π^{2}.

#### Answer: 25

At the highest point, the vertical component of velocity is zero.

So, at the highest point, we have:

velocity = *v* = *u*cos*θ*

Centripetal force on the particle = mv2r

⇒ mv2r = mu2cos2θr

At the highest point*,* we have : mg=mv2r

Here,* r *is the radius of curvature of the curve at the point.

⇒ r = u2cos2θg

#### Answer: 26

Let *u* be the initial velocity and *v* be the velocity at the point where it makes an angle θ2with the horizontal component.

It is given that the horizontal component remains unchanged.

Therefore, we get:

v cos θ2 = u cosθ

⇒ v = ucosθcosθ2 …(i)

mgcosθ2 = mv2r ..(ii)

⇒ r = v2gcosθ2

On substituting the value of *v* from equation (i), we get:

r = u2cos2θgcos2θ2

#### Answer: 27

Given:

Radius of the room = *R*

Mass of the block = *m*

(a) Normal reaction by the wall on the block = *N *= mv2R

(b) Force of frictional by the wall = μN = μmv2R

(c) Let *a*_{t} be the tangential acceleration of the block.

From figure, we get:

-μmv2R = mat

⇒ at = -μv2R

(d) On using a = dvdt = vdvds,

we get : vdvds = μv2 R

⇒ ds = -Rμdvv ,

Integrating both side,

we get : s = -RμInv + cAt,

s = 0, v = v0So, c = RμInv0

⇒ s = -RμInvv0

⇒ vv0 = e-μsR

⇒ v = v0e – μsR For one rotation,

we have:

s = 2πr

∴ v = v0e – 2πμ

#### Exercise : Solution of Questions on page Number : 116

#### Answer: 28

Let the mass of the particle be *m.*

Radius of the path* = R *

Angular velocity = *ω*

Force experienced by the particle = *mω*^{2}*R*

The component of force *mR*ω^{2} along the line AB (making an angle with the radius) provides the necessary force to the particle to move along AB.

∴ mω2R cosθ = ma

⇒ a = ω2Rcosθ

Let the time taken by the particle to reach the point *B* be *t.*

On using equation of motion,

we get: L = ut + 12at2

⇒ L = 12ω2R cosθt2

⇒ t2 = 2Lω2R cosθ

⇒ t = 2Lω2R cosθ

#### Answer: 29

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 = mv2r = 0.1 × 10050 = 15 = 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 = mv2rcosθ …(i)

μN = mv2rsinθ …(ii)

On using i and ii,

we get : μmv2rcosθ = mv2rsinθ

⇒ μ = tanθ

⇒ θ = tan-1 (0.58)≈30°

#### Answer: 30

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 – ma-mω2R = 0 …(i)

T + 2ma-2mω2R = 0 …(ii)

On subtracting eq. (i) by eq. (ii),

we get : 3ma = mω2R

⇒ a = ω2R3

Substituting the value of *a *in eq. (*i*), we get:

T = mω2R3 + mω2R

⇒ T = 43 mω2R