WBCHSE Class 11 Physics Circular Motion Long Answer Questions

Circular Motion Long Answer Type Questions

Question 1. Why is a centripetal force necessary for a uniform circular motion?
Answer:

According to Newton’s first law of motion, if there is no external force acting on a body, the body remains at rest or moves with uniform velocity. So, to rotate a body along a circular path, its inertia has to be overcome.

For this reason, an external force has to be applied to the body. This external force acts towards the centre of the circular path radially and is called the centripetal force. In the absence of this force, uniform circular motion is not possible.

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Question 2. Centripetal force is called a real force, but centrifugal force is called a pseudo force. Give reasons in support of this statement.
Answer:

The force that arises due to a mutual action-reaction process is called a real force. We know that in the absence of any external force, a moving body continues to move with uniform velocity.

  • So, with respect to any inertial frame of reference, when a body moves along a circular path, at every moment its direction of motion changes.
  • This change in direction of motion is possible when an external force acts on the body. This external force is the centripetal force.
  • As centripetal force is an external force acting on a rotating body, it is a real force.
  • On the other hand, in an accelerated or rotating frame of reference, i.e., in a non-inertial frame of reference, a body is acted upon by a force whose direction is opposite to the direction of the centripetal acceleration of the frame.
  • Since this force is not generated due to a mutual action-reaction process between different bodies, it is a pseudo force.
  • This force has no existence in any inertial frame of reference. This force, which arises due to an acceleration of the reference frame, can be felt only in the non-inertial frame of reference.
  • In a rotating frame of reference, the pseudo force, which acts on a body and is equal but opposite to the centripetal force is called centrifugal force.

Question 3. What should be the length of a day on the earth when a body has no apparent weight at the equatorial region? (Radius of the earth = 6400 km) Or, What should be the time period of rotation of the earth about its own axis so that a person on the equator feels weightless? (Equatorial radius = 6400 km ,g = 9.8m · s-2)
Answer:

Weight mg of a body acts towards the centre of the earth. If the radius of the earth is R, at the equatorial region, the centrifugal force experienced by the body is mω²R, where, ω is the angular velocity of the earth about its own axis.

When the centrifugal force acting on the body becomes equal and opposite to the weight of the body, the apparent weight of the body becomes zero. In that case,

mg = \(m \omega^2 R\)

or, \(\omega=\sqrt{\frac{g}{R}}\) = time period of rotation of the earth (T)

= \(\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{R}{g}}=2 \pi \times \sqrt{\frac{6400 \times 10^3}{9.8}}=5077.6 \mathrm{~s}\)

= \(1 \mathrm{~h} 24.6 \mathrm{~min}\).

Question 4. What is the maximum speed with which a car can move over a convex bridge?
Answer:

To move over a convex bridge, the car requires a centripetal force. This centripetal force is provided by the weight of the car.

Let us consider that when the car moves with a certain velocity v, the weight of the car is just sufficient to provide that necessary centripetal force; but if the car moves with a velocity greater than v, the car loses its contact with the bridge.

Let the mass of the car be m and the radius of the circular path be r. So, the condition for the car to not lose contact with the surface of the bridge is \(\frac{m v^2}{r}\) ≤ m g or,  v ≤ \(\sqrt{g r}\)

So, a car can move with a maximum speed of √gr over a convex bridge of radius of curvature r such that it does not lose contact with the bridge.

Question 5. Radius of a curved path is r and the coefficient of friction between the wheel of a car and that path is μ. What should be the maximum speed with which the car can take a turn without skidding in that curved path?
Answer:

Weight of the car = mg; so normal force offered by the path = mg.

Hence, limiting frictional force = μmg.

[μ = coefficient of friction between the wheel of the car and the path]

The limiting frictional force provides the centripetal force necessary to prevent skidding.

So, if the maximum speed of the car is v, then the condition for no skidding in that path is, \(\frac{m v^2}{r}=\mu m g \quad \text { or, } \quad v=\sqrt{\mu r g} \text {. }\)

Question 6. While taking a turn, is it possible for a cyclist to lean | at an angle of 45° with the vertical?
Answer:

We know that when a cyclist takes a turn along a circular path of radius r with velocity v, he leans at an angle of θ with the vertical.

In this case, \(\tan \theta=\frac{v^2}{r g}\)

Again, normal force on the cycle = mg, where m is the mass of the cycle along with the cyclist. So, if the coefficient of friction is μ, then limiting frictional force = μmg. This limiting frictional force provides the necessary centripetal force.

Hence, \(\frac{m v^2}{r}=\mu m g \quad \text { or, } \quad \mu=\frac{v^2}{r g}\)…(2)

From the equations (1) and (2) we get, tanθ = μ.

It is given that θ = 45°; so, μ = tan45° = 1.

But the coefficient of friction between the wheel and the road can never be 1; actually, it is less than 1.

Hence, the cycle will skid when the cyclist leans at an angle of 45° with the vertical.

Question 7. A stone is tied to the end of a string. When the string is whirled in a circular path, the string can never be kept horizontal. Explain the reason.
Answer:

When a stone tied to the end of a string is whirled along a horizontal circular path, the horizontal component of the tension in the string provides the necessary centripetal force and its vertical component balances the weight of the stone.

If the stone is rotated along a horizontal circular path by keeping the string horizontal, there would be no vertical component of the tension (T) in the string. In that case, the weight of the body (mg) cannot be balanced.

Circular Motion A Stone Is Tied To The End Of String

As a result, the stone would dip downwards leaving the horizontal circular path until the vertical component of T would be just sufficient to balance the weight of the stone. Hence, during rotation of the stone along a horizontal circular path, the string can never be kept horizontal.

Question 8. When a body is hung from a string, it does not snap, But when the same mass is set into rotation along a horizontal path at high speed holding the other end of the string, the string snaps. What is the reason?
Answer:

If the mass of the body is m, then in the first case, tension in the string, T’ = mg ….(1)

Let the length of the thread be l, linear velocity of the body be v, and tension in the thread be T, when the body is rotated in the horizontal plane.

Circular Motion Body Is Hung From A String

Required centripetal force for revolution, \(\frac{m v^2}{l \sin \theta}=T \sin \theta\)…(2)

Also, \(m g=T \cos \theta\)…(3)

From equations (1) and (3) we get, \(T^{\prime}=T \cos \theta\)

∴ 0 ≤ θ <\(\frac{\pi}{2}\)

Circular Motion Centripetal Force For Revolutions

∴ 0< cosθ ≤1

∴ \(T^{\prime}<T\)

So tension in the thread is greater in the second case than that in the first case, and hence, in the first case though the thread does not snap, in the second case the thread may snap.

Question 9. Two identical trains are running in opposite directions over two tracks along the equator at equal speed. Will both the trains exert the same force on the tracks?
Answer:

Both trains will not exert the same force on the tracks. The train, running towards the east from the west, possesses a greater angular velocity than that of the earth with respect to the axis of rotation of the earth. This is because the earth is also rotating from the west to the east with respect to the same axis.

  • As a result, the outward centrifugal force, which is directly proportional to aω², will increase. Hence, the apparent weight of the train will be reduced. The apparent weight is the difference of the real weight and the centrifugal force on the train.
  • Consequently, the train will exert comparatively less force on the tracks. On the other hand, the train running towards the west from the east will have an increase in its apparent weight, and hence, will exert a comparatively greater force on the tracks.

Question 10. A hollow cylinder of radius r is rotating about its own vertical axis. Both ends of the cylinder are open. A piece of stone of mass m remains fixed on the inner side of the cylinder. What should be the minimum velocity of rotation of the cylinder so that the stone remains fixed on the surface of the cylinder without falling down? Coefficient of static friction between the wall of the cylinder and the stone = μ, acceleration due to gravity = g. Also, prove that this velocity is independent of the mass of the stone.
Answer:

The forces acting on the piece of stone are

  1. Weight of the stone, W = mg
  2. Limiting frictional force, F = μR

[ft = coefficient of friction, R = normal force of the wall,]

Circular Motion A Hollow Cylinder Of Radius

Minimum velocity of rotation = v;

The radius of the cylinder = r.

In equilibrium (i.e., when the stone remains fixed to the wall),

W = mg = F or, mg = μR

Again, R = \(\frac{m v^2}{r}\)

∴ mg = \(\mu \cdot \frac{m v^2}{r}\text { or, } v^2=\frac{g r}{\mu}\text { or, } v=\sqrt{\frac{g r}{\mu}}\)

The above expression is independent of mass.

Question 11. When a moving bus takes a right turn, the passengers inside the bus seem to lean towards the left Explain why.
Answer:

When a bus takes a turn, the centrifugal force felt by the passengers is directed opposite to the direction of the turn. Hence, when a bus takes a right turn, the passengers inside the bus seem to lean towards the left.

Question 12. A small glass marble is put on a smooth gramophone disc. When the disc starts rotating, the marble flies off the disc. Explain why.
Answer:

When the disc starts rotating, the marble flies off because the frictional force between the surface of the gramophone disc and the marble cannot provide the necessary centripetal force required by the marble for its revolution.

Question 13. In a circus show, a motorcycle rider inside a ‘death- -well’ can revolve on the erect wall without falling down. What is the reason behind it?
Answer:

While revolving on the erect wall inside the well, the rider derives the necessary centripetal force for revolving in a circular path from the horizontal component of the normal force. Due to the presence of this normal force, an upward frictional force is generated which balances the combined weight of the motorcycle and the cyclist. For this reason, the rider does not fall down.

Question 14. Why should a signboard mentioning the safe maximum speed be cited before the bend on a horizontal road or railway track?
Answer:

If the speed of a vehicle is more than the maximum safe limit written on the board, the frictional force acting on the tyres of the vehicle by the road cannot provide the necessary centripetal force for its turning, and hence, there is a chance of skidding.

To avoid an accident, the driver of the vehicle should know the maximum speed before he reaches the bend. The signboard provides him with that knowledge.

Question 15. When a motor car travels on a convex road the passengers inside feel lighter. Why?
Answer:

When a motor car travels on a convex road, the resultant of the weight (W) of any passenger inside it and the upward normal force (R) on the surface of the car supplies the necessary centripetal force to the passenger to travel along the convex road (i.e., circular path),

i.e., \(W-R=\frac{m v^2}{r}\),

where, m = mass of the passenger,

v = linear velocity of the car

and r = radius of the circular path

or, R = \(W-\frac{m v^2}{r}\)…(1)

Due to this normal force R, the passenger feels this weight. M=Now, according to equation (1), since R < W, the passenger feels lighter.

Question 16. A funnel is rotating around its vertical axis with a constant frequency ν rev/s. A small cube of mass m is placed on the inside wall of the funnel carefully. The wall of the funnel makes an angle θ with the horizontal. If μ is the coefficient of friction between the funnel and the cube, r is the distance between the centre of mass of the cube and the rotational axis, then find the maximum and minimum value of v for which the cube remains static with respect to the funnel.

Circular Motion A Funnel Is Rotating Around Its Vertical Axis

Answer:

Frequency of rotation of the funnel = ν rev/s.

Circular Motion Rotational Of The Funnel

The centripetal force required for the cube of mass m to rotate around the circular path of radius r is,

Fr = mω²r = m(2πv)²r = 4π²ν²mr

Let the block slide down the wall of the funnel when the funnel is at rest. Suppose, v1 is the minimum value of frequency for which the block remains static. At that time, the direction of the frictional force will be upward along the walls of the funnel.

Balancing the forces on the cube, we get

mg =(ncosθ + fsinθ) [n = normal reaction, f = limiting value of the static friction]

= \(n \cos \theta+\mu n \sin \theta\)

= \(n(\cos \theta+\mu \sin \theta)\)…(1)

and \(m \omega_1^2 r=n \sin \theta-f \cos \theta=n \sin \theta-\mu n \cos \theta\)

= \(n(\sin \theta-\mu \cos \theta)\)….(2)

From equations (1) and (2) we have, \(\frac{\omega_1^2 r}{g}=\frac{\sin \theta-\mu \cos \theta}{\cos \theta+\mu \sin \theta}\)

or, \(\left(2 \pi \nu_1\right)^2=\frac{g}{r}\left(\frac{\sin \theta-\mu \cos \theta}{\cos \theta+\mu \sin \theta}\right)\)

or, \(\nu_1=\frac{1}{2 \pi} \sqrt{\frac{g}{r}\left(\frac{\sin \theta-\mu \cos \theta}{\cos \theta+\mu \sin \theta}\right)}\)

Now, let v2 be the maximum value of the frequency of rotation for which the cube remains static with respect to the funnel. Then the direction of the frictional forces will be downward along the walls of the funnel.

Circular Motion Rotational Of The Funnel

Balancing the forces on the cube, we get mg = \(n \cos \theta-f \sin \theta=n(\cos \theta-\mu \sin \theta)\)…(3)

and \(m \omega_2^2 r=n(\sin \theta+\mu \cos \theta)\)…(4)

From equations (3) and (4) we get, \(\frac{\omega_2^2 r}{g}=\frac{\sin \theta+\mu \cos \theta}{\cos \theta-\mu \sin \theta}\)

or, \(\nu_2=\frac{1}{2 \pi} \sqrt{\frac{g}{r}\left(\frac{\sin \theta+\mu \cos \theta}{\cos \theta-\mu \sin \theta}\right)}\)

Question 17. Why are the bends of a road or of a railway track banked?
Answer:

At the bend of horizontal roads or railway tracks, frictional force provides the necessary centripetal force for a car or a train to take a turn.

  • If the radius of curvature at the bend is low and if the frictional force is not large enough, the speed of the car or train at the bend has to be lowered to avoid skidding.
  • To avoid this difficulty, instead of depending on friction alone, the roads or railway tracks are banked to provide additional centripetal force. The necessary centripetal force is supplied by the horizontal component of the normal force of the plane of the road or railway tracks.
  • Hence, the outer side of the road or railway tracks is slightly elevated with respect to the inner side.

Question 18. Why do we feel lighter and heavier at the highest and lowest points of a Ferris wheel? Suppose, the Ferris wheel is revolving with a constant angular velocity.
Answer:

At the highest point of the wheel, we have mg – n = \(\frac{m v^2}{r}\)

where m = mass of the passenger, v = linear velocity of the passenger at the rotating Ferris wheel, r = radius of curvature of the Ferris wheel and n is the normal reaction.

∴ n = \(m g-\frac{m v^2}{r}\)

So, we feel lighter at the highest point.

At the lowest point of the trajectory \(n-m g=\frac{m v^2}{r} \quad \text { or, } n=m g+\frac{m v^2}{r}\)

Hence, we feel heavier at the lowest point of the Ferris wheel.

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