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

`\alpha =10.93radian/sec^2`

Step-by-step 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* \alpha * 13.8`

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

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