Physics: Mechanical Energy

label Physics
account_circle Unassigned
schedule 1 Day
account_balance_wallet $5

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

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

Did you know? You can earn $20 for every friend you invite to Studypool!
Click here to
Refer a Friend
...
Oct 16th, 2017
...
Oct 16th, 2017
Oct 17th, 2017
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer