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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 radI 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

Answer»

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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