## Related questions with answers

A torque of 50.0 N-m is applied to a grinding wheel (I = 20.0 kg-m²) for 20 s. (a) If it starts from rest, what is the angular velocity of the grinding wheel after the torque is removed? (b) Through what angle does the wheel move through while the torque is applied?

Solution

Verified$\textbf{a)}$ we know that :

$\tau = I \alpha$

where :

- $I$ = 20 kg$\cdot$m$^2$ is the moment of inertia of the flywheel .
- $\tau$ = 50 N$\cdot$m is the torque.

$\textbf{Plugging}$ known information into the torque equation we get:

$\begin{align*} \tau &= I \alpha \\ \alpha&= \dfrac{\tau }{I}\\ &= \dfrac{50}{20}\\ &=2.5 \end{align*}$

$\boxed{\alpha = 2.5\ \; \mathrm{rad/ s^2}}$

From $\textbf{the kinematics of the rotational motion}$ we knw that :

$\omega_{f} = \omega_{i} + \alpha t$

Where:

- $\omega_{i}$ is the initial angular velocity .
- $\omega_{f}$ is the final angular velocity .
- $\alpha$ is the angular acceleration .
- $t$ is the time
- $\Delta \theta$. is the angle.

From $\textbf{givens}$ we know that : $\alpha = 2.5\ \; \mathrm{rad/s}$ , because it starts from rest $\omega_{i} = 0\ \; \mathrm{rad/s}$ and $t = 20\ \; \mathrm{s}$ .

$\textbf{Plugging}$ known information into our equation we get :

$\begin{align*} \omega_{f}& = \omega_{i} + \alpha t \\ &= 0 + 2.5 \times 20 \\ &=50 \end{align*}$

$\boxed{\omega_{f} = 50\ \; \mathrm{rad/s}}$

## Create a free account to view solutions

## Create a free account to view solutions

## Recommended textbook solutions

#### Physics for Scientists and Engineers: A Strategic Approach with Modern Physics

4th Edition•ISBN: 9780133942651 (8 more)Randall D. Knight#### Mathematical Methods in the Physical Sciences

3rd Edition•ISBN: 9780471198260 (1 more)Mary L. Boas#### Fundamentals of Physics

10th Edition•ISBN: 9781118230718 (3 more)David Halliday, Jearl Walker, Robert Resnick#### University Physics, Volume 1

1st Edition•ISBN: 9781938168277Jeff Sanny, Samuel J Ling, William Moebbs## More related questions

1/4

1/7