Question

A wheel with a 0.10-m radius is rotating at 35 rev/s. It then slows uniformly to 15 rev/s over a 3.0-s interval.

a. What is the angular acceleration of a point on the wheel?
b. how to do and explain will give lifesaver!

Answer 1

Answer:

The angular acceleration of a point on the wheel is $41.89\ rad/s^2$ and it is decelerating.

Explanation:

It is given that,

Radius of the wheel, r = 0.1 m

Initial angular velocity of the wheel, $\omega_i=35\ rev/s=219.91\ rad/s$

Final angular velocity of the wheel, $\omega_f=15\ rev/s=94.24\ rad/s$

Time, t = 3 s

We need to find the angular acceleration of a point on the wheel. It is given by the rate of change of angular velocity divided by time taken. It is given by :

$\alpha =\dfrac{\omega_f-\omega_i}{t}$

$\alpha =\dfrac{(94.24-219.91)\ rad/s}{3\ s}$

$\alpha =-41.89\ rad/s^2$

So, the angular acceleration of a point on the wheel is $41.89\ rad/s^2$ and it is decelerating. Hence, this is the required solution.