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

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Answer 1 · 2017-12-01T04:12:43

Torque: $\qquad \tau = I\alpha = (80\pi)/3\quad kg.m^2. (rad)/s^2$

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: $\qquad I = 1/2 MR^2 = 10 \quad kg.m^2$

Angular Acceleration: $\qquad \alpha = (\Delta\omega)/(\Deltat) = (8\pi)/3\quad (rad)/s^2$

Torque: $\qquad \tau = I\alpha = (80\pi)/3\quad kg.m^2. (rad)/s^2$

The rate of rotation of a solid disk with a radius of 2 m2m2 m and mass of 5 kg5kg5 kg constantly changes from 12 Hz12Hz12 Hz to 16 Hz16Hz16 Hz. If the change in rotational frequency occurs over 3 s3s3 s, what torque was applied to the disk?