On the Wheel of Fortune game show, a solid wheel (of radius 7.4 m and mass 10 kg

Question

On the Wheel of Fortune game show, a solid wheel (of radius 7.4 m and mass 10 kg) is given an initial counterclockwise angular velocity of +1.11 rad/s. It then smoothly slows down and stops after rotating through 3/4 of a turn.

On the Wheel of Fortune game show, a solid wheel (of radius 7.4 m and mass 10 kg) is given an initial counterclockwise angular velocity of +1.11 rad/s. It then smoothly slows down and stops after rotating through 3/4 of a turn. (a) Find the frictional torque that acts to stop the wheel. N · m (b) Assuming that the torque is unchanged, but the mass of the wheel is doubled and its radius is halved. How will this affect the angle through which it rotates before coming to rest? Increase? Decrease? Stay the same?

Explanation / Answer

1.)

Torque =(moment of inertia)(angular acceleration)
Moment of inertia of a solid wheel = 1/2 m R^2 = (0.5)(10kg)(7.4m)^2 = 273.8 kg m^2
Angular acceleration = (w^2 -w0^2)/2 theta

That's the formula we use when the time of motion is NOT given.

Angular acceleration = [0^2 -(1.11rad)^2]/ 2(theta)
Comment: w(final angular speed)=0 (b/c the wheel stops)
theta(the angle through which the wheel turns) = 3/4 turn = (3 pi)/2 radians
Now the torque = (273.8kg m^2)(-0.130 rad/s^2) = -35.79 Nm

2.)

If mass is doubled and radius is halved,then

Moment of inertia of a solid wheel = 1/2 m R^2 = (0.5)(20kg)(3.7m)^2 = 136.9 kg m^2

T = Ia

-35.79 Nm = 136.9 kg m^2 * a

a = -0.261

Angular acceleration = [0^2 -(1.11rad)^2]/ 2(theta)

-0.261 = [0^2 -(1.11rad)^2]/ 2(theta)

theta = 2.36rad Ans

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