Question

A constant torque of 23 N-m is applied to a shaft that is rotating with a CW angular velocity of 22.6 rad/s. Determine the time required to reverse the direction of motion and achieve a CCW rotation of 22.6 rad/s.

Answer 1

Answer:

2.78 s

Explanation:

We are given that

Torque,$\tau=23 Nm$

Initial angular velocity,$\omega_0=22.6 rad/s$

Final angular velocity,$\omega=-22.6 rad/s$

Mass,m=2 kg

R=0.53 m

r=0.27 m

Moment of inertia of system=$I=2m(R^2+r^2)$

$I=2\times 2((0.53)^2+(0.27)^2}=1.4152 kgm^2$

Angular acceleration,$\alpha=-\frac{\tau}{I}=-\frac{23}{1.4152}=-16.25rad/s^2$

$\alpha=\frac{\omega-\omega_0}{t}$

$t=\frac{\omega-\omega_0}{\alpha}$

$t=\frac{-22.6-22.6}{-16.25}=2.78 s$