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 bya 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

Answer 1

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 ω = 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]*ω +m(L/2)²*ω

Initial angular momentum = Final angular momentum

mv (L/2) = [(1/3) ML²]*ω + m(L/2)²* ω

{ [(1/3) ML² ] +m(L/2)² } ω= mv(L/2)

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

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

≈0.3 rad/s

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(b) The mud does NOT make a significantcontribution.