Question

(a) What is the magnitude of the tangential acceleration of a bug on the rim of a 12.5-in.-diameter disk if the disk accelerates uniformly from rest to an angular speed of 75.0 rev/min in 3.80 s

Answer 1

Answer:

Explanation:

Given:

d=12.5in=0.3175m

r=d/2=0.3175/2=0.15875m

ωf=75rev/min=7.85rad/s

t=3.80s

The angular acceleration

$\alpha = \frac{\omega _f - \omega _i}{t}=\frac{7.85-0}{3.80}=2.07rad/s^2$

Tangential acceleration

$\alpha =r\alpha =0.15875\times  2.07=0.33m/s^2$