Question

A 4.50-kg wheel that is 34.5 cm in diametet rotates through an angle of 13.8 rad as it slows down uniformly from 22.0 rad/s to 13.5 rad/s. What is the magnitude of the angular acceleration of the wheel? A) 10.9 rad/s^2 B) 0.616 rad/s^2 C) 22.5 rad/s^2 D) 111 rad/s^2 E) 5.45 rad/s^2

Answer 1

Answer:

$\alpha =10.93radian/sec^2$

Explanation:

We have given given the final angular velocity $\omega _{final}=13.5rad/sec$

And $\omega _{initial}=22rad/sec$

Displacement $\Theta =13.8radian$

We have to find the angular acceleration $\alpha$

According to law of motion $\omega _{final}^2=\omega _{initial}^2+2\alpha \Theta$

So $13.5^2=22^2+2\times \alpha \times 13.8$

$\alpha =-10.93radian/sec^2$

In question we have tell about magnitude only so $\alpha =10.93radian/sec^2$