Question

A cylinder rotating about its axis with a constant angular acceleration of 1.6 rad/s^2 starts from rest at t=0. At the instant when it has turned through 0.40 radian, find the magnitude of the tangential velocity at point on the rim (radius=0.15 m)?

Answer 1

Using 3rd rotational kinematic equation:

w^2 = w0^2 + 2**

= angular displacement = 0.40 rad

= angular acceleration = 1.6 rad/sec^2

w0 = Initial angular velocity = 0 rad/sec

w = final angular velocity = ?

w = sqrt (w0^2 + 2**)

w = sqrt (0^2 + 2*1.6*0.40) = 1.1314 rad/sec

Now at this time tangential velocity will be:

V = w*R

R = radius of cylinder = 0.15 m

So,

V = 1.1314*0.15

V = 0.17 m/s = tangential velocity