A 98.1-kg horizontal circular platform rotates freely with no friction about its

Question

A 98.1-kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.79 rad/s. A monkey drops a 9.17-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 1.73 m.

Explanation / Answer

Moment of inertia of the circular disc I1  =1/2 m r2=1/2X.98.1X.1.73X.1.73 = 146.8 Kg m2

When monkey drops bunch of banana, the angular velocity of the disc gets reduced to conserve angular momentum because its moment of inertia will be increased . there is no externel torque acting on the system.

Now moment of inertia of disc along with banana I2 = 146.8 +9.17 x ( 1.73x4/5)2 =164.36 kg.m2. Applying the law of conservation of angular momentum we have

    I1 X W1 =I2X W2      ( I1 =146.8 W1 = 1.79 rad/sec ,I2 =164.36 kg m2 W2 = M I of the disc with banana)

146.8 x 1.79 =164.36 x W2 ; W2 =1.6 radian /sec

When monkey falls on the disc , there is further increase in the M. I. of the disc, if it is I3 then

I3= 164.36 + 21.3 X 1.73X1.73 = 228.10 kg m2

Applying again law of conservation of angular mometum , I2 X W2 = I3 X W

W =(164.36 X1.6 ) / 228.10 = 1.15 radian/sec

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