This set of Physics Multiple Choice Questions & Answers (MCQs) focuses on “Dynamics of Rotational Motion about a Fixed Axis”.

1. A rigid body is rotating about an axis. One force F_{1} acts on the body such that its vector passes through the axis of rotation. Another force F_{2} acts on it such that it is perpendicular to the axis of rotation and at a point 5cm from the axis. This force F_{2} is perpendicular to the radius vector at its point of application. Find the net torque on the body. Let F_{1} = 10N & F_{2}= 5N.

a) 0

b) 10.25Nm

c) 0.25Nm

d) 10Nm

View Answer

Explanation: The force F

_{1}passes through the axis of rotation, so it will not produce a torque.

The force F

_{2}is perpendicular to axis and radius, so it will provide a torque = r*F

_{2}about the axis of rotation, where ‘r’ is the distance of point of application of force F

_{2}from the axis of rotation.

Therefore, torque = r* F

_{2}= 0.05 * 5 Nm

= 0.25Nm.

2. A ring is rotating about a diameter. The radius = 5cm & mass of ring = 1kg. A force is applied on the ring such that it is perpendicular to the axis and vector **AB** as shown in the figure. The magnitude of force is 10N. Find the work done by the torque, when the ring rotates by 90°.

a) 0.25π J

b) 0.15π J

c) 0.2π J

d) 0

View Answer

Explanation: Work done is given by Tθ, where T is the torque & θ is the angular displacement = π/2 rad. From the given figure, OA = 3cm & OB = 5cm,

therefore AB

^{2}= 5

^{2}– 3

^{2}= 25 – 9 = 16 OR AB = 4cm.

Torque due to force F will be AB*F = 0.04*10 = 0.4Nm.

∴ Work = Tθ = 0.4*π/2 = 0.2π J.

3. A ring of radius 7cm is rotating about the central axis perpendicular to its plane. A force acts on it, tangentially, such that it does a work of 10J in a complete rotation. Find the value of that force. `

a) 0.5/7π N

b) 35π N

c) 500/7π N

d) 0.7 N

View Answer

Explanation: Let the force be ‘F’. Work done will be = Tθ,

where T is the torque & is the angular displacement = 2π rad.

Work = Tθ = 10

∴ T(2π) = 10

∴ T = 5/π Nm.

Also, T = r*F

∴ 5/π = 0.07*F

∴ F = 500/7π N.

4. A disc of radius 10cm is rotating about the central axis perpendicular to its plane. A force of 5N acts on it tangentially. The disc was initially at rest. Calculate the value of power supplied by the force when the disc has rotated by 30°.The mass of the disc is 2kg.

a) 1.28W

b) 1.8W

c) 3.9W

d) 2.63W

View Answer

Explanation: Work done = Tθ, where T is torque & θ is angular displacement.

T = r*F = 0.1*5 = 0.5Nm

& θ = 30° = π/6 rad.

∴ Work = 0.5*π/6 = π/12 J.

Moment of inertia about spinning axis = MR

^{2}/2 = 2*0.01/2 = 0.01kgm

^{2}.

Using, T = Ia, where ‘I’ is moment of inertia & ‘a’ is angular acceleration, we get:

a = T/I

∴ a = 0.5/0.01 = 50 rad/s

^{2}.

Now, θ = w

_{0}t + (1/2)at

^{2}to calculate the time for 30° rotation, where w

_{o}is initial angular velocity.

∴ π/6 = 0 + 0.5*50*t

^{2}.

∴ t = √(π/150) = 0.145 s.

∴ Power = Work/time = (π/12) / (0.145) = 1.8 W.

5. A rod is rotating about one end. If a force F_{1} acting on the other end produces a torque T & supplies power P, find the value of force F_{2} that will produce the same amount of power when it acts at the midpoint of the rod. Assume that all forces are perpendicular to the axis of rotation & axis of rod. The rod starts from rest.

a) = F_{1}

b) > F_{1}

c) < F_{1}

d) No force acting at any other point will produce the same power

View Answer

Explanation: Power is defined as the work done per unit time. For the same power, the amount of rotation should be the same in a given time. For the th angular acceleration should be the same in both the cases & therefore the torque should be the same.

Thus, F

_{2}* I/2 = F

_{1}* I.

F

_{2}= F

_{1}*2 OR F

_{2}> F

_{1}

**Sanfoundry Global Education & Learning Series – Physics – Class 11**.

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