Question

A 89.3 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.77 rad/s . A monkey drops a 8.83 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 21.1 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 1.69 m .

Answer 1

Answer:

The angular velocity of the platform is 1.11 rad/s

Explanation:

The initial angular momentum is equal to:

$L_{o} =\frac{1}{2} m_{o} R_{o} ^{2} w_{o}$

Where

m₀ = 89.3 kg

R₀ = 1.69 m

w₀ = 1.77 rad/s

When the moment is conserved, then:

L₀ = L₁

$L_{1} =\frac{1}{2} m_{o} R_{o} ^{2} w_{1} +(m_{1} R_{1}^{2} )w_{1}$

Where

m₁ = 8.83 kg

R₁ = (4/5)R₀

The monkey drops vertically, the moment is conserved, then:

L₀ = L₂

$L_{2} =\frac{1}{2} m_{o} R_{o} ^{2} w_{2} +(m_{1} R_{1}^{2} )w_{2}+(m_{2} R_{o}^{2} )w_{2}\\\frac{1}{2} m_{o} R_{o} ^{2} w_{o}=\frac{1}{2} m_{o} R_{o} ^{2} w_{2}+(m_{1}(4/5)^{2} R_{o} ^{2})w_{2}+(m_{2} R_{o}^{2} )w_{2}\\w_{2}=\frac{m_{o}w_{o}}{m_{o}+(32/25)m_{1}+2m_{2}}$

Replacing:

$w_{2}=\frac{89.3*1.77}{89.3+(32/25)*8.83+(2*21.1)} =1.11rad/s$

Answer 2

Answer:

$\omega_{2} = 1.107\,\frac{rad}{s}$

Explanation:

Let assume that circular platform is a solid cylinder. Given the absence of external forces, the situation can be analyzed by applying the Principle of Angular Momentum, which states that:

$I_{1}\cdot \omega_{1} = I_{2}\cdot \omega_{2}$

The initial moment of inertia is:

$I_{1} = \frac{1}{2}\cdot (89.3\,kg)\cdot (1.69\,m)^{2}$

$I_{1} = 127.525\,kg\cdot m^{2}$

Likewise, the final moment of inertia is:

$I_{2} = 127.525\,kg\cdot m^{2} + (8.83\,kg)\cdot (1.352\,m)^{2} + (21.1\,kg)\cdot (1.69\,m)^{2}$

$I_{2} = 203.929\,kg\cdot m^{2}$

The final angular speed is:

$\omega_{2} = \frac{I_{1}}{I_{2}} \cdot \omega_{1}$

$\omega_{2} = \frac{127.525\,kg\cdot m^{2}}{203.929\,kg\cdot m^{2}} \cdot (1.77\,\frac{rad}{s} )$

$\omega_{2} = 1.107\,\frac{rad}{s}$