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 +1.12 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. vector tau = 112.3 Incorrect: Your answer is incorrect. 74.9 was also incorrect N · m

i need a quick answer!!!!

Answer 1

angular Torque =(moment of inertia)(angular acceleration)
Moment of inertia of a solid wheel = 1/2 m R^2 = (0.5)(10kg)(7.5m)^2 = 281.25 kg m^2
Angular acceleration = (w^2 -w0^2)/2 theta {newtons laws of motion }

Angular acceleration = [0^2 -(1.12rad)^2]/ 2(theta)
Comment: w(final angular speed)=0 rad/sec
theta(the angle through which the wheel turns) = 3/4 turn = (3 pi)/2 radians

angular acceleration = = -0.133 Rad/sec^2
Now the torque = (281.25 kg m^2)(-0.133 rad/s^2) = -37.433 Nm Negative sign indicates the torque is slowing down the wheel.