Answer
(a) $F = 67.6~N$
(b) $F = 62.9~N$
(c) $t = 3.27~s$
Work Step by Step
(a) We can find the required magnitude of angular acceleration.
$\alpha = \frac{\omega_f-0}{t}$
$\alpha = \frac{(2.00~rev/s)(2\pi~rad/rev)}{9.00~s}$
$\alpha = 1.396~rad/s^2$
We can find the required force $F$ on the end of the crank handle.
$\sum \tau = I~\alpha$
$(F)(0.500~m)-(F_f)(0.260~m) - 6.50~N~m = \frac{1}{2}MR^2\alpha$
$(F)(0.500~m) = \frac{1}{2}MR^2\alpha+(F_N)(\mu_k)(0.260~m) + 6.50~N~m$
$(F)(0.500~m) = \frac{1}{2}(50.0~kg)(0.260~m)^2(1.396~rad/s^2)+(160~N)(0.60)(0.260~m) + 6.50~N~m$
$(F)(0.500~m) = 33.82~N~m$
$F = 67.6~N$
(b) Since the disk should turn at a constant angular speed, the angular acceleration should be zero. Then the net torque should be zero. Therefore, the torque from the applied force should be equal in magnitude to the sum of the torque from the ax's friction and the torque from the friction in the axle.
$F~(0.500~m) = (F_N)(\mu_k)(0.260~m) + 6.50~N~m$
$F = \frac{(160~N)(0.60)(0.260~m) + 6.50~N~m}{0.500~m}$
$F = 62.9~N$
(c) We can find the magnitude of angular deceleration when the torque from the axle is slowing down the grindstone.
$6.50~N~m = I\alpha$
$\alpha = \frac{6.50~N~m}{\frac{1}{2}(50.0~kg)(0.260~m)^2}$
$\alpha = 3.846~rad/s^2$
We can find the time it takes for the grindstone to come to a stop.
$t = \frac{\Delta \omega}{\alpha}$
$t = \frac{0-(2.00~rev/s)(2\pi~rad/rev)}{-3.846~rad/s^2}$
$t = 3.27~s$
