A 91.1-kg horizontal circular platform rotates freely with no friction about its

Question

A 91.1-kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.69 rad/s. A monkey drops a 9.63-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 22.1 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 1.89 m.?

A 91.1-kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.69 rad/s. A monkey drops a 9.63-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 22.1 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 1.89 m. Number rad/s Tools x 102

Explanation / Answer

The moment of inertia of disk is given by

I = ½ * m * r2 = ½ * 91.1 * (1.89)2 = 162.71 kg.m2

Also, the initial angular momentum is

Li = 162.71 * 1.69 = 274.98

For the bananas, r = 4/5 * 1.89 = 1.512 m
therefore,

I = 9.63 * (1.512)2 = 22.015 kg.m2

For the monkey, I = 22.1 * (1.89)2 = 78.94 kg.m2

Thus ,Total moment of inertia is

I = 162.71 + 22.015 + 78.94 = 263.665 kg.2

Now, we have

initial angular momentum = final angular momentum

The final angular momentum = 263.665 * ?
therefore,

263.665 * ? = 274.98
? = 274.98 ÷ 263.665

? = 1.0429 rad/s

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