# The rate of rotation of a solid disk with a radius of 2 m and mass of 5 kg constantly changes from 12 Hz to 16 Hz. If the change in rotational frequency occurs over 3 s, what torque was applied to the disk?

Dec 1, 2017

Torque: $\setminus q \quad \setminus \tau = I \setminus \alpha = \frac{80 \setminus \pi}{3} \setminus \quad k g . {m}^{2.} \frac{r a d}{s} ^ 2$

#### Explanation:

Given:
\omega_i = (2\pi \quad rad)(12 Hz) = 24\pi \quad (rad)/s;
\omega_f = (2\pi\quad rad)(16 Hz) = 32\pi \quad (rad)/s;

\Delta\omega = 8\pi\quad (rad)/s; \qquad \Deltat = 3s; \qquad M = 5kg; \qquad R = 2m

Moment-of-Inertia - Solid Disc: $\setminus q \quad I = \frac{1}{2} M {R}^{2} = 10 \setminus \quad k g . {m}^{2}$

Angular Acceleration: $\setminus q \quad \setminus \alpha = \frac{\setminus \Delta \setminus \omega}{\setminus \Delta t} = \frac{8 \setminus \pi}{3} \setminus \quad \frac{r a d}{s} ^ 2$

Torque: $\setminus q \quad \setminus \tau = I \setminus \alpha = \frac{80 \setminus \pi}{3} \setminus \quad k g . {m}^{2.} \frac{r a d}{s} ^ 2$