# 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.

Category: QuestionsA 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.
Editor">Editor Staff asked 1 year ago

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.

I have been asked this question in exam.

This intriguing question comes from Kinematics of Rotational Motion about a Fixed Axis in portion System of Particles and Rotational Motion of Physics – Class 11
Select the correct answer from above options
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Editor">Editor Staff answered 1 year ago

Correct choice is (a) 19/3 rad

For 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.

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