Question

A bug is on the rim of a disk of diameter 9 in that moves from rest to an angular speed of 76 rev/min in 6.5 s. what is the tangential acceleration?

Answer 1

Let's convert the final angular speed of the disk into rad/s:
$\omega_f = 76 \frac{rev}{min} \cdot \frac{2 \pi rad/rev}{60 s/min}=7.96 rad/s$
while the initial angular speed is zero.

So, the angular acceleration of the disk is
$\alpha = \frac{\omega_f - \omega_i }{t} = \frac{7.96 rad/s}{6.5 s}=1.22 rad/s^2$

The radius of the disk is half the diameter: $r= d/2 = 9 m/2 =4.5 m$

And so, the tangential acceleration is the angular acceleration times the radius:
$a_t = \alpha r =(1.22 rad/s^2)(4.5 m)=5.49 m/s^2$