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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

Respuesta :

Answer:

[tex]\alpha =10.93radian/sec^2[/tex]

Explanation:

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

And [tex]\omega _{initial}=22rad/sec[/tex]

Displacement [tex]\Theta =13.8radian[/tex]

We have to find the angular acceleration [tex]\alpha[/tex]

According to law of motion [tex]\omega _{final}^2=\omega _{initial}^2+2\alpha \Theta[/tex]

So [tex]13.5^2=22^2+2\times \alpha \times 13.8[/tex]

[tex]\alpha =-10.93radian/sec^2[/tex]

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