This set of Physics online quiz focuses on “Kinematics of Rotational Motion about a Fixed Axis”.

1. The rate of change of angular speed is constant. What will be the expression for change in angular speed after time dt, if angular acceleration is ‘a’? The initial angular speed is ‘w’.

a) a*dt

b) w – a*dt

c) w + a*dt

d) There is no change

View Answer

Explanation: The equation of rotational motion is given by: w = w

_{0}+ a*dt, where dt is the time interval. Therefore, the change in angular speed = w – w

_{0}= a*dt.

2. The angular speed of a body is 2rpm. What should be the angular acceleration for the angular speed to double up in 5s?

a) -2π/75 rad/s^{2}

b) -π/75 rad/s^{2}

c) π/75 rad/s^{2}

d) 2π/75 rad/s^{2}

View Answer

Explanation: Initial angular speed = 2rpm = π/15 rad/s. Final angular speed = 2π/15 rad/s.

Using, w = w

_{0}+ a*dt, we get:

2π/15 = π/15 + 5a.

∴ π/15 = 5*a

OR a = π/75 rad/s

^{2}.

3. A body has a moment of inertia = 5kgm^{2} about an axis. What should be the torque for increasing its angular speed from 0 to 10rad/s in 4s? Assume that the body rotates purely about that axis.

a) 25Nm

b) 2.5Nm

c) 12.5Nm

d) 1.25Nm

View Answer

Explanation: w = w

_{0}+ a*dt OR 10 = 0 + a(4)

∴ a = 2.5 rad/s

^{2}.

Torque = Ia = 5*2.5 = 12.5Nm,

where ‘I’ is the moment of inertia.

4. A body is in pure rotational motion about an axis. Its angular acceleration is given by: a = 2t. What is the angular distance covered by the body from t=2s to t=3s? It starts from rest at t=0.

a) 19/3 rad

b) 0 rad

c) 2/3 rad

d) 5/3 rad

View Answer

Explanation: Let w be the angular velocity & a be the angular acceleration. Given: dw/dt = a = 2t. On integrating both sides, we get:

w = t

^{2}(no integration constant because w=0 at t=0).

Now, dθ/dt = w = t

^{2}

∴ θ = t

^{3}/3.

To find angular distance covered from t=2 to t=3, we can subtract distance covered in 2s from distance covered in 3s.

∴ θ covered = 3

^{3}/3 – 2

^{3}/3 = 19/3 rad.

5. A circular disc is rotating about a central axis perpendicular to its diameter with a speed of 5rad/s. If a constant force of 2N is applied tangentially on it, in how much time will it come to a stop? Radius of disc = 1m & mass = 1kg.

a) 0.257s

b) 5s

c) 1.25s

d) 2.5s

View Answer

Explanation: Let r be the radius, I be the moment of inertia, a be the angular acceleration & F be the force applied. Moment of inertia of disc about given axis = MR

^{2}/2 = 0.5 kgm

^{2}.

Torque = r X F = 2Nm.

But, torque is also = Ia.

∴ a = 2/0.5 = 4 rads

^{2}.

Using, w = w

_{0}+ a*dt, we get:

0 = 5 – 4dt, where dt is the time interval.

∴ dt = 5/4 = 1.25s.

6. A body is under an angular deceleration of 5rad/s^{2}. Its initial speed is 3rad/s. How much angular distance ‘θ’ will it cover before coming to rest?

a) 0.8 rad

b) 0.9 rad

c) 9 rad

d) 2 rad

View Answer

Explanation: Let w be the angular velocity & a be the angular acceleration.

Using, w

^{2}= w

_{0}

^{2}+ 2aθ, we get:

0 = 9 – 2*5*θ

Or 9 = 10θ

∴ θ = 0.9 rad.

7. A ring of diameter 1m is rotating about a central axis perpendicular to its diameter. It is rotating with a speed of 10rad/s. What force should be applied on it tangentially to stop it in exactly 3 rotations? Let the mass of the ring be 2kg.

a) 10 N

b) 25/3π N

c) 50/3π N

d) 12.5/3 N

View Answer

Explanation: Let w be the angular velocity & a be the angular acceleration.

Moment of inertia of the ring = MR

^{2}

= 2*0.25 = 0.5kgm

^{2}.

Angular distance = 3 rotations = 3*2π rad = 6π rad.

Using, w

^{2}= w

_{0}

^{2}+ 2aθ, we get:

0 = 100 – 2a*6π.

a = 100/12π.

Torque = Ia = rF

∴ F = Ia/r = 0.5*100/(12π*0.5) = 25/3π N.

8. The angular acceleration of a body = 3t^{2}. What will be its speed at t=10s, if initial speed = 10rad/s?

a) 1010 rad/s

b) 300 rad/s

c) 1310 rad/s

d) 1000rad/s

View Answer

Explanation: Angular acceleration a = dw/dt = 3t

^{2}. On integrating, we get:

w = t

^{3}+ c.

At t=0, w=10,

∴ c = 10.

Therefore, w = t

^{3}+ 10.

Therefore, w at t=10

= 1000+10 = 1010 rad/s.

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