Question

A wheel on a game show is given an initial angular speed of 1.12 rad/s . It comes to rest after rotating through 0.75 of a turn.

Part A. Find the average torque exerted on the wheel given that it is a disk of radius 0.61 m and mass 5.4 kg.
Part B. If the mass of the wheel is doubled and its radius is halved, will the angle through which it rotates before coming to rest increase, decrease, or stay the same?
(Assume that the average torque exerted on the wheel is unchanged.)
will increase? will decrease? will stay same?

Answer 1

Answer:

A. T = 0.134 Nm

B. will increase

Explanation:

$\Delta \theta$ = 0.75 turn = 0.75 * 2π rad/turn = 4.71 rad

We can use the following equation of motion to find out the angular acceleration of the wheel:

$\omega^2 - \omega_0^2 = 2\alpha\Delta \theta$

where $\omega = 0 m/s$ is the final angular velocity of the wheel when it stops, $\omega_0 = 1.12rad/s$ is the initial angular velocity of the wheel, $\alpha$ is the deceleration, and $\Delta \theta = 4.71 rad$ is the angle traveled

$0 - 1.12^2 = 2\alpha*4.71$

$\alpha = 1.12^2 / (2*4.71) = 0.133 rad/s^2$

A. The moment of inertia of the solid disk:

$I = mR^2/2$

Where m = 5.4 kg is the disk mass and R = 0.61 m is the radius of the disk.

$I = 5.4*0.61^2/2 = 1.00467 kgm^2$

So the torque exerted is

T = αI = 0.133*1.00467 = 0.134 Nm

B. mass is doubled = 5.4 *2 = 10.8 kg

radius is halved = 0.61 / 2 = 0.305 m

The new moment of inertia of the solid disk:

$I = mR^2/2$

Where m = 10.8 kg is the disk mass and R = 0.305 m is the radius of the disk.

$I = 10.8*0.305^2/2 = 0.502335 kgm^2$

If torque stays the same, the new angular acceleration is

$\alpha = T/I = 0.134 / 0.502335 = 0.267 rad/s^2$

Since 0.267 > 0.133, the angular acceleration will increase