A solid wood door 1.00 m wide and 2.00 m high is hinged along one side and has a

Question

A solid wood door 1.00 m wide and 2.00 m high is hinged along one side and has a total mass of 45.0kg . Initially open and at rest, the door is struck at its center by a handful of sticky mud with mass 0.700 kg, traveling perpendicular to the door at 12.0m/s just before impact

A)Find the final angular speed of the door.
answer in rad/s

B)Does the mud make a significant contribution to the moment of inertia?
Yes or No

so stuuck please your help will be appreciated clearly my prophs not that good at explaining

Explanation / Answer

Massof door (M) = 45.0 kg

  Mass of mud (m) = 0.700 kg

Let width of the door be L = 1 m

Speedof the mud (v) = 12 m/s

Let   = angular velocity

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Consider solid wood door and mud assystem then there are no exdternal torques

acting on the systemthen angular momentum is conserved. then

         Initial angular momentum =   angular momentumof mud   =   m v (L/2)

      FinalAngular momentum is = ANGULAR MOMENTUM OF DOOR +ANGULAR MOENTUM OF MUD

                                                = [(1/3)ML2]* +m(L/2)2*

          Initial angular momentum = Final angular momentum

                                 mv (L/2) = [(1/3) ML2]* + m(L/2)2*

      { [(1/3) ML2 ] +m(L/2)2 }    = mv(L/2)

                                                  = [mv (L/2)] /    { [(1/3) ML2 ] +m(L/2)2 }

Substitute all the above values solve for =0.276 rad/s

                                                              0.3 rad/s

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