Asked by Z
A horizontal 810-N merry-go-round of radius 1.70 m is started from rest by a constant horizontal force of 55 N applied tangentially to the merry-go-round. Find the kinetic energy of the merry-go-round after 2.0 s. (Assume it is a solid cylinder.)
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Answered by
drwls
The kinetic energy is (1/2) I w^2 , where I is the moment of inertia, which is (1/2) M R^2 for a solid cylinder. Therefore KE = (1/2) M R^2 w^2
w is the angular velocity in radians per second. To determine its value after 2.0 seconds, you will need to use the equation of motion
T = I*dw/dt
where T is the torque, which in this case is 55 * 1.70 = 93.5 N m/s. You will need to compute dw/dt, the rate of change of angular velocity.
The value of w at 2.0 s is
w (@t=2) = dw/dt * 2.0 radians per second.
Now put all that information together and do the calculation.
w is the angular velocity in radians per second. To determine its value after 2.0 seconds, you will need to use the equation of motion
T = I*dw/dt
where T is the torque, which in this case is 55 * 1.70 = 93.5 N m/s. You will need to compute dw/dt, the rate of change of angular velocity.
The value of w at 2.0 s is
w (@t=2) = dw/dt * 2.0 radians per second.
Now put all that information together and do the calculation.
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