Question

What is the tangential acceleration of a bug on the rim of a 10.0 in. diameter disk if the disk moves from rest to an angular speed of 75 revolutions per minute in 4.0 s?

Answer 1

What is the tangential acceleration of a bug on the rim of a 10.0 in. diameter disk if the disk moves from rest to an angular speed of 75 revolutions per minute in 4.0 s?

**Solution:**

$d = 10.0$ in $= 0.254\mathrm{m}$ –diameter of the disk;

$v = 75$ revolutions per minute $= \frac{75}{60} \cdot 2\pi = 7.85\frac{\mathrm{rad}}{\mathrm{s}}$ –angular speed of the disk;

$t = 4\mathrm{s}$ – time that disk need to move from rest to an angular speed.

Angular acceleration of the disk is:

$$a_{\text{ang}} = \frac{v}{t} = \frac{7.85 \frac{\mathrm{rad}}{\mathrm{s}}}{4\mathrm{s}} = 1.96 \frac{\mathrm{rad}}{\mathrm{s}^2}$$

Tangential acceleration of the disk:

$$a_{\text{ang}} = \frac{a_{\text{tan}}}{r} \Rightarrow a_{\text{tan}} = a_{\text{ang}} \cdot r = a_{\text{ang}} \cdot \frac{d}{2} = 1.96 \frac{\mathrm{rad}}{\mathrm{s}^2} \cdot \frac{0.254\mathrm{m}}{2} = 0.25 \frac{\mathrm{m}}{\mathrm{s}^2}$$

**Answer:** Tangential acceleration of the disk is $0.25 \frac{\mathrm{m}}{\mathrm{s}^2}$.