A 80.5-kg horizontal circular platform rotates freely with no friction about its

Question

A 80.5-kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.57 rad/s. A monkey drops a 9.51-kg bunch of bananas vertically onto the platform. They hit the platform at 4/5 of its radius from the center, adhere to it there, and continue to rotate with it. Then the monkey, with a mass of 21.3 kg, drops vertically to the edge of the platform, grasps it, and continues to rotate with the platform. Find the angular velocity of the platform with its load. Model the platform as a disk of radius 1.57 m.

Explanation / Answer

The initial angular momentum is, L = I = ½ m R ² ,

where, m = 80.5 kg, R = 1.57 m, = 1.57 rad / s

Further, dropping the bunch of bananas onto the platform doesn’t exert a sideward force along the
plane of rotation of the platform, so that angular momentum is conserved.
L = L , where, L = ( ½ m R ² ) + ( m R ² )

where, m = 9.51 kg, R = ( / ) R,

when the monkey drops himself vertically to the edge of the platform and remains there, the total angular
momentum is still conserved … L = L

where, L = ( ½ m R ² ) + ( m R ² ) + ( m R ² )

so that,
½ m R ² = ( ½ m R ² ) + [ m ( / ) ² R ² ] + ( m R ² )
m R ² = [ m + 2 m ( ¹ / ) + 2 m ] R ²
m = [ m + 2 m ( ¹ / ) + 2 m ]
= m / [ m + ( ³² / ) m + 2 m ]

where, m + ( ³² / ) m + 2 m = 80.5 kg + ( ³² / ) ( 9.51 kg ) + 2 ( 21.3 kg ) = 135.27 kg

= ( 80.5 kg ) ( 1.57 rad / s ) / 135.27 kg = 0.934 rad /s

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