Question

a wheel on a game show is given an initial angular speed of 1.22 rad/s. it comes to rest after rotating through 0.75 of a turn. find the average torque exerted on the wheel given that it is a disk of radius 0.71 m and mass 6.4 kg.
a. -0.16 n.m
b. -0.69 n.m
c. -0.25 n.m
d. -0.44 n.m
e. -0.31 n.m

Explanation:

Step1: Convert turns to radians

The wheel rotates through 0.75 turns. Since 1 turn = $2\pi$ radians, the angular displacement $\theta=0.75\times2\pi = 1.5\pi$ radians. The initial angular speed $\omega_0 = 1.22$ rad/s and the final angular speed $\omega = 0$ rad/s.

Step2: Find the angular acceleration $\alpha$

Use the equation $\omega^{2}=\omega_{0}^{2}+2\alpha\theta$. Rearranging for $\alpha$ gives $\alpha=\frac{\omega^{2}-\omega_{0}^{2}}{2\theta}$. Substituting the values: $\alpha=\frac{0 - (1.22)^{2}}{2\times1.5\pi}=\frac{- 1.4884}{3\pi}\approx - 0.158$ rad/$s^{2}$.

Step3: Calculate the moment of inertia $I$

For a disk, the moment of inertia $I=\frac{1}{2}mr^{2}$, where $m = 6.4$ kg and $r = 0.71$ m. So $I=\frac{1}{2}\times6.4\times(0.71)^{2}=3.2\times0.5041 = 1.61312$ kg$\cdot$m$^{2}$.

Step4: Find the average torque $\tau$

Use the equation $\tau=I\alpha$. Substituting the values of $I$ and $\alpha$: $\tau=1.61312\times(- 0.158)\approx - 0.25$ N$\cdot$m.

Answer:

C. -0.25 N.m