Question

A body is moving down into a well through a rope passing over a fixed pulley of radius $20\text{cm}$. Assume that there is no slipping between the rope and pulley. Calculate the angular velocity and angular acceleration of the pulley at an instant when the body is going down at a speed of $40\text{cm/s}$ and has an acceleration of $4{\text{m/s}}^2$?

• A. $\omega =2\text{rad/s},\alpha =20{\text{rad/s}}^2$
• B. $\omega =2\text{rad/s},\alpha =40{\text{rad/s}}^2$
• C. $\omega =1\text{rad/s},\alpha =20{\text{rad/s}}^2$
• D. $\omega =1\text{rad/s},\alpha =40{\text{rad/s}}^2$

Answer 1

The correct option is A $\omega =2\text{rad/s},\alpha =20{\text{rad/s}}^2$

Since the rope does not slip on the pulley, the linear speed $v$ of the rim is same as the speed of the body.
Given, $r=20\text{cm}$ (Radius of pulley)
$v=40\text{cm/s}$ (Speed of body)
$a=4{\text{m/s}}^2$ (Acceleration of body)
The angular velocity of pulley is $\omega =\frac{v}{r}=\frac{40\text{cm/s}}{20\text{cm}}$
$\omega =2\text{rad/s}$
The angular acceleration of the pulley is
$\alpha =\frac{a}{r}=\frac{4{\text{m/s}}^2}{20\text{cm}}=\frac{400{\text{cm/s}}^2}{20\text{cm}}$
or $\alpha =20{\text{rad/s}}^2$