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A child of mass 48.8 kg sits on the edge of a merry-go-round with radius 2.9 m and moment of inertia 320.118 kgm2. The merrygo-round rotates with an angular velocity of 3 rad/s. The child then walks towards the center of the merry-go-round and stops at a distance 1.16 m from the center. Now what is the angular velocity of the merry-go-round?

1 Answers

Hello!

For this problem you must first consult the Conservation of Angular Momentum. The force the child needs to hold on will be equal to the centripetal force.

Centripetal Force is given by F = m(v^2)/r = m(w^2)r

This gives a force of F = (48.8kg)(3rad/sec)^2(2.9m) = 1273.68 N

Once the child moves towards the center, The angular momentum must be conserved. Li = Lf

The moment of inertia for the child is given by Ic = mr^2 and the merry-go-round has inertia Im = 320.118 kgm^2

By equating the initial and final angular momentum you will get the equation (Ic1 + Im)w1 = (Ic2 + Im)w2

So we will solve for w2 which is the angular velocity of the merry-go-round after the child has moved towards the center.

w2 = w1(mr1^2 + Im) / (mr2^2 + Im)

w2 = 3[(48.8kg * (2.9m)^2 + 320.118kgm^2] / (48.8kg * (1.16m)^2 + 320.118kgm^2)

w2 = 5.68085 rad/s

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