Question

A 91.1 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.73 rad/s . A monkey drops a 9.41 kg bunch of bananas vertically onto the platform. They hit the platform at 45 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 1.61 m .

Answer 1

Answer:

$\omega_f = 1.08 rad/s$

Explanation:

As we know that there is no external torque on the system of platform, banana and the monkey

so the angular momentum of the system will remains conserved

so here we can say

$L_i = L_f$

$I_o\omega = (I_o + m_1r_1^2 + m_2r_2^2)\omega_f$

here we know that

$I_o = \frac{1}{2}mR^2$

$I_o = \frac{1}{2}(91.1)(1.61)^2 = 118.1 kg m^2$

$m_1 = 9.41 kg$

$r_1 = \frac{4}{5}(1.61) = 1.29 m$

$m_2 = 21.1 kg$

Now from above equation

$118.1(1.73) = (118.1 + 9.41(1.29^2) + 21.1(1.61^2))\omega_f$

$118.1(1.73) = 188.45\omega_f$

$\omega_f = 1.08 rad/s$