Question

A merry -go-round rotates at the rate of 0.4rev / s with an 78 kg man standing at a point 1.5 m from the axis of rotation. What is the new angular speed when the man walks to a point m from the center? Consider the merry-go-round is a solid 11 kg cylinder of radius of 1,5 m. Answer in units of rad/s. 024 (part 2 of 2) 10.0 points What is the change in kinetic energy due to this movement ? Answer in units of

Answer 1

1) Final angular velocity is 5.22 rad/s

2) Change in kinetic energy: 639.8 J

Explanation:

1)

In absence of external forces, the angular momentum of the merry-go-round must be conserved.

So we can write:

$L_1 = L_2\\I_1\omega_1 = I_2 \omega_2$

where

$I_1$ is the initial moment of inertia

$I_2$ is the final moment of inertia

$\omega_1$ is the initial angular velocity

$\omega_2$ is the final angular velocity

At the beginning, we have:

$\omega_1=0.4 rev/s \cdot 2\pi = 2.51 rad/s$

The moment of inertia is the sum of the moment of inertia of the cylinder + the moment of inertia of the man, therefore:

$I_1 = \frac{1}{2}MR^2+mr^2=\frac{1}{2}(11)(1.5)^2+(78)(1.5)^2=187.9 kgm^2$

The man later walks to a point 1 m from the centre, so the final moment of inertia is:

$I_2 = \frac{1}{2}MR^2+mr^2=\frac{1}{2}(11)(1.5)^2+(78)(1)^2=90.4 kgm^2$

Therefore, the final angular velocity is

$\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(187.9)(2.51)}{90.4}=5.22 rad/s$

2)

The angular velocity of the merry-go-round rotating is given by

$K=\frac{1}{2}I\omega^2$

In the  first situation, we have

$I_1 = 187.9 kg m^2\\\omega=2.51 rad/s$

Therefore

$K_1 = \frac{1}{2}(187.9)(2.51)^2=591.8 J$

In the second situation, we have

$I_2= 90.4 kg m^2\\\omega_2=5.22 rad/s$

Therefore

$K_2=\frac{1}{2}(90.4)(5.22)^2=1231.6 J$

So the change in kinetic energy is

$\Delta K = K_2 - K_1 = 1231.6-591.8=639.8 J$

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