Physics: Mechanical Energy

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A horizontal 810-N merry-go-round of radius 1.70 m is started from rest by a constant horizontal force of 45 N applied tangentially to the merry-go-round. Find the kinetic energy of the merry-go-round after 4.0 s. 

(Assume it is a solid cylinder. Also assume the force is applied at the outside edge.) 

Oct 16th, 2017

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Mass of the merry go round =810/9.8 =82.65 kg






  =2.17 m/s

Linear Kinetic energy =1/2 m v^2

                                   =1/2 * 82.65* 2.17^2

                                    =194.22  Joules

let us find angular or rotational kinetic energy

   I =1/2 mR^2

      =1/2 *82.65 * 1.70^2

      =119.42 kg m^2

Torque =F.r


            =76.5 =Ia

angular velocity w=T *t/I


                              =2.56 rad /s

Rotational kinetic energy =1/2 I w^2

                                         =1/2 *119.42 *2.56^2

                                         =391.3 J

T.K.E =391.3+194.22

          =585.52 J


Please let me know if you need any clarification. I'm always happy to answer your questions.
Jul 5th, 2015

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