Question

n electric motor rotating a workshop grinding wheel at 1.10 102 rev/min is switched off. Assume the wheel has a constant negative angular acceleration of magnitude 1.92 rad/s2. (a) How long does it take the grinding wheel to stop?

Answer 1

Time taken to stop = 6 sec

Step-by-step explanation:

Given:

Rotation of wheel = 1.10 × 10² rev/min

Final velocity (v) = 0

Angular acceleration (a) = - 1.92 rad/s²

Find:

Time taken to stop

Computation:

Initial velocity (u) = (1.10 × 10² × 2π rad) / 60 sec

Initial velocity (u) = 11.52 rad / sec

We know that,

V = U +at

t = (v-u)a

t = (0 - 11.52) / (-1.92)

Time taken to stop = 6 sec