The combination of an applied force and a frictional force produces a constant t

Question

The combination of an applied force and a frictional force produces a constant torque of 33 N · m on a wheel rotating about a fixed axis. The applied force acts for 4.9 s, during which time the angular speed of the wheel increases from 0 to 11 rad/s. What is the moment of inertia of the wheel? Answer in units of kg · m2. 026 (part 2 of 3) 10.0 points The applied force is then removed, and the wheel comes to rest in 71 s. What is the frictional torque? Answer in units of N · m. 027 (part 3 of 3) 10.0 points How many revolutions does the wheel make during the entire 76 s interval? Answer in units of rev.

Explanation / Answer

Part 1) Combined torque, Tc = I = I/t

=> I = Tct1/ = 33 * 4.9 / (11 - 0) = 14.7 kg-m2

Part 2) Frictional torque, Tf = I/t2 = 14.7 * 11 / 71 = 2.28 N-m

Part 3) For the first 4.9 s, 1 = 1t12/2 = (11 / 4.9) * 4.92 / 2 = 26.95 rad

For the next 71 s, 2 = 2t22/2 = (11 / 71) * 712 / 2 = 390.5 rad

Total revolutions, = (390.5 + 26.95) / 2 = 66.44 rev

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