Question

A spinning wheel initially has an angular velocity of 50rad/s east: 20s later its angular velocity is 50rad/s west. If the angular acceleration is constant, what are (a) the magnitude and direction of angular acceleration, (b) the angular displacement over 20 s and (c) the angular speed?

Answer 1

(a) initial angular velocity of wheel, $\omega_0=50\hat{i}$ rad/s

final angular velocity, $\omega=-50\hat{i}$ rad/s

time taken , t = 20sec

we know,

angular acceleration = change in angular velocity/time taken

= $\frac{-50\hat{i}-50\hat{i}}{20}$

= $-5\hat{i}$ rad/s²

hence, magnitude of angular acceleration is 5 rad/s² and it acts along west direction.

(b) use formula, $\theta=\omega_0t+\frac{1}{2}\alpha t^2$

where, $\theta$ denotes angular displacement and $\alpha$ denotes angular acceleration.

so, $\theta$ = 50 × 20 + 1/2 × (-5) × 20²

= 1000 - 1000 = 0

hence, angular displacement = 0

(c) angular speed = angular distance/time

angular distance = 50 × 20 + 1/2 × 5 × 20² = 1000 + 1000 = 2000 rad
[ just use magnitude of angular acceleration]

so, angular speed = 2000/20 = 100 rad/s