Question

A constant torque of 19.2 N · m is applied to
a grindstone for which the moment of inertia
is 0.11 kg · m2.
Find the angular speed after the grindstone
has made 11.2 rev. Assume the grindstone
starts from rest.
Answer in units of rad/s

Answer 1

\(\tau = I *\alpha \) \(\rightarrow \alpha = \dfrac{\tau}{I}= \dfrac{19.2}{0.11}= 174.5 ~~rad/s^2\)
and you have formula \(\theta = \theta_0+ \omega_0 *t +\dfrac{1}{2}\alpha t^2\)where \(\theta = 11.2*2\pi\)
so, time t = 0.898 s
replace this time to formula \(\omega = \omega_0 +\alpha t\) you will have the answer is 156.7 rad/s

Answer 2

Thank you! :)