Homework Help Forum: physics

Posted by Z on Sunday, November 1, 2009 at 2:12am.

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.)

• physics - drwls, Sunday, November 1, 2009 at 6:08am

  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.

  
  

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