A playground carousel (“merry-go-round”) is free to rotate frictionlessly in the horizontal plane (and air resistance is negligible). Without riders, the carousel has a moment of inertia of 152 kg·m2. But there is a single rider, initially standing 1.85 m from the axis of rotation, as the carousel turns at an angular speed of 0.640 rad/s. Then the person moves to another location, 0.75 m from the axis, and the angular speed is then 0.973 rad/s. Find the person’s mass.

Question

A playground carousel (“merry-go-round”) is free to rotate frictionlessly in the horizontal plane (and air resistance is negligible). Without riders, the carousel has a moment of inertia of 152 kg·m2. But there is a single rider, initially standing 1.85 m from the axis of rotation, as the carousel turns at an angular speed of 0.640 rad/s. Then the person moves to another location, 0.75 m from the axis, and the angular speed is then 0.973 rad/s. Find the person’s mass.

Answer 1

To solve this problem, we can use the principles of conservation of angular momentum. Angular momentum is given by the equation:

L = I * ω

where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

In the initial state, the person is standing 1.85 m from the axis of rotation, and the angular speed is 0.640 rad/s. So, we can write the equation for the initial angular momentum:

L_initial = I * ω_initial

In the final state, the person moves to a location 0.75 m from the axis of rotation, and the angular speed is 0.973 rad/s:

L_final = I * ω_final

Since angular momentum is conserved, we can equate the initial and final angular momentum:

L_initial = L_final

I * ω_initial = I * ω_final

Simplifying the equation, we can cancel out the moment of inertia:

ω_initial = ω_final

Now, we can solve for the person's mass. The formula for moment of inertia of a point mass is given by:

I = m * r^2

where I is the moment of inertia, m is the mass, and r is the distance from the axis of rotation.

In the initial state, the person is standing 1.85 m from the axis of rotation:

I_initial = m * r_initial^2

In the final state, the person moves to a location 0.75 m from the axis of rotation:

I_final = m * r_final^2

Since the moment of inertia is the same in both states, we can set I_initial equal to I_final:

m * r_initial^2 = m * r_final^2

Now we can solve for the person's mass:

m = (r_final^2 * I) / r_initial^2

Substituting the given values, we have:

m = (0.75^2 * 152 kg·m^2) / 1.85^2

m = 90.953 kg

Therefore, the person's mass is approximately 90.953 kg.