Question

A 10 KgKg wheel that is 40 cmcm in diameter rotates through an angle of 19 radrad as it slows down uniformly from 16 rad/secrad/sec to 26 rad/secrad/sec . What is the magnitud of the angular acceleration of the wheel?

Answer 1

Answer:

The magnitude of angular acceleration of the wheel = 0.26 $\frac{rad}{s^{2} }$

Explanation:

Mass m = 10 kg

Diameter = 0.4 m

Initial speed $\omega_{1} = 26 \ \frac{rad}{sec}$

Final speed $\omega_{2} = 16 \ \frac{rad}{sec}$

Angle $\theta$ = 19 rad

The value of final speed is given by

$\omega _{2} = \omega _{1} + 2 \alpha \theta$

Put all the values in above equation we get

16 = 26 + 2 ($\alpha$) (19)

$\alpha$ = - 0.26 $\frac{rad}{s^{2} }$

Therefore the magnitude of angular acceleration of the wheel = 0.26 $\frac{rad}{s^{2} }$