On the Wheel of Fortune game show, a solid wheel (of radius 7.5 m and mass 10 kg

Question

On the Wheel of Fortune game show, a solid wheel (of radius 7.5 m and mass 10 kg) is given an initial counterclockwise angular velocity of +1.17 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. torque = 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 stay the same

Explanation / Answer

W^2 = W0^2 + 2*alpha*theta

0 = 1.17^2 + 2*alpha*1.5pi

=> alpha = -0.145 rad/s^2

Torque = I*alpha ( I = MR^2 in case of ring)

=> torque = 10*7.5^2*0.145 = - 81.56 N.m

b) if mass is halved , and radius is doubled the I will be double => alpha will be halved if torque is constant , so theta will increase

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