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.) 

Jul 5th, 2015

Thank you for the opportunity to help you with your question!

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

F=ma

a=F/m

v/t=F/m 

v=Ft/m

v=45*4/82.65

  =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

            =45*1.70

            =76.5 =Ia

angular velocity w=T *t/I

                              =76.5*4/119.42

                              =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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