On the Wheel of Fortune game show, a solid wheel (of radius 7 m and mass 10 kg)

Question

On the Wheel of Fortune game show, a solid wheel (of radius 7 m and mass 10 kg) is given an initial counterclockwise angular velocity of +1.22 rad/s. It then comes to rest after rotating through 3/4 of a turn.

(a) Find the frictional torque that acts to stop the wheel.
tau vec = ___ N · m

(b) Assuming that the torque is unchanged, but the mass of the wheel is halved and its radius is doubled. How will this affect the angle through which it rotates before coming to rest? (Make sure you can reason this with proportionalities.)

Increase, decrease, or stay the same?

Explanation / Answer

^2= 2*a*

a = R

=1.22

= (3/4)*2

solving we have,

= 0.02256 rad/s2

Net torque T = I*

I = MR2/2

T = 5.527 N-m

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