# Angular Acceleration of game show wheel

1. Mar 15, 2005

### tsdemers

A wheel on a game show is given an initial angular speed of 1.22 rad/s. It comes to rest after rotating through 3/4 of a turn. Find the average torque exerted on the wheel given that it is a disk of radius 0.71m and a mass 6.4kg.

I'm lost. I don't really know where to begin. I know the answer is going to be 0.25Nm but I just need some help getting on the right track. Any suggestions?

Last edited: Mar 15, 2005
2. Mar 15, 2005

### Andrew Mason

Use conservation of energy

Work done = $\tau \Delta \theta$ = Energy available = $\frac{1}{2}I\omega^2$

AM

3. Mar 15, 2005

### xanthym

From the problem statement:
{Disk Mass} = M = (6.4 kg)
{Disk Radius} = R = (0.71 m)
{Disk Moment of Inertia} = I = (1/2)*M*R2 = (1/2)*(6.4)*(0.71)2 = (1.613 kg*m2)
{Initial Disk Angular Speed} = ω0 = (1.22 rad/s)
{Final Disk Angular Speed} = ωf = (0.0 rad/s)
{Disk Angular Rotation} = θ = (3/4 Turn) = (4.712 rad)
{Torque Applied to Disk} = τ

From Conservation of Rotational Kinetic Energy:
{Rotational Work} = τ*θ = {Final Rotational Kinetic Energy} - {Initial Rotational Kinetic Energy}
::: ⇒ τ*θ = (1/2)*I*(ωf)2 - (1/2)*I*(ω0)2
::: ⇒ τ*θ = (1/2)*I*{(ωf)2 - (ω0)2}
::: ⇒ τ*(4.712) = (-1.2)
::: ⇒ |τ| = (0.255 N*m)

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4. Mar 15, 2005

### tsdemers

Thank you

Thank you for the help. It's greatly appreciated