A small 4.00 kg brick is released from rest 3.00 m above a horizontal seesaw on

Question

A small 4.00 kg brick is released from rest 3.00 m above a horizontal seesaw on a fulcrum at its center, as shown in the figure below .
The distance between the fulcrum and the falling brick is 1.6m.

https://s3.amazonaws.com/answer-board-image/5ec028df3348d4e7fe858f0e5ae01310.jpg

A) Find the angular momentum of this brick about a horizontal axis through the fulcrum and perpendicular to the plane of the figure the instant the brick is released.

B)Find the angular momentum of this brick about a horizontal axis through the fulcrum and perpendicular to the plane of the figure the instant before it strikes the seesaw.

Explanation / Answer

a) the distance of brick ( before falling ) from the fulcrum is (32+1.62) = 11.56=3.4 m

thus, angular momentum in given case = mvr , but since v=0 m/s, so , L = mvr = 4 x 0 x 3.4 = 0 kg m2/s

b) in falling 3.00 meters , vel gained by brick = v= (2gh)=(2x10x3)=7.75 m/s

so, angular momentum = L = mvr = 4.00 x 7.75 x 1.6 kg m2/s = 49.6 kg m2/s

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