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 (rev)/(min) \cdot (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 = (\omega_f - \omega_i )/(t) = (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`