A merry-go-round rotates at the rate of 0.16 rad/s with a(n) 43.0 kg man standing at a point 2.4 m from the axis of rotation.

What is the new angular speed when the man walks to a point 0.64 m from the center? Assume that the merry-go-round is a solid 5.99 ×102 kg cylinder with a radius of 3.87 m.
Answer in units of rad/s.

To find the new angular speed of the merry-go-round when the man walks to a point closer to the center, we need to apply the law of conservation of angular momentum.

The angular momentum L of an object is given by the equation L = Iω, where I is the moment of inertia and ω is the angular speed.

Before the man walks, the total angular momentum of the system is given by the sum of the man's angular momentum and the merry-go-round's angular momentum.

The moment of inertia I of the merry-go-round can be calculated using the formula for a solid cylinder: I = 0.5 * m * r^2, where m is the mass of the merry-go-round and r is its radius.

The angular momentum of the man is calculated by multiplying his moment of inertia (which can be approximated as a point mass) by his angular speed: L man = m man * ω man.

The total initial angular momentum is the sum of these two components: L initial = L man + L merry-go-round.

When the man walks closer to the center, the moment of inertia of the system changes because the distribution of mass changes. The new moment of inertia I' can be calculated using the same formula as before, but with the new distance r' from the axis of rotation.

Since angular momentum is conserved, the initial angular momentum is equal to the final angular momentum: L initial = L final.

To find the final angular speed ω', we rearrange the equation and solve for ω' = L final / I'.

Substituting the expressions for the initial angular momentum and the new moment of inertia, we get:

L initial = L man + L merry-go-round
L final = m man * ω' + I' * ω'

Setting these two expressions equal to each other, we have:

m man * ω man + I * ω = m man * ω' + I' * ω'

Since we want to find ω', we solve for it:

m man * ω' = m man * ω man + I * ω - I' * ω'

Simplifying the equation, we have:

ω' = (m man * ω man + I * ω - I' * ω') / m man

Now we can substitute the given values into the equation to calculate the new angular speed:

m man = 43.0 kg (mass of the man)
ω man = 0.16 rad/s (initial angular speed of the man)
I = 0.5 * 5.99 × 10^2 kg * (3.87 m)^2 (moment of inertia of the merry-go-round)
r' = 0.64 m (new distance of the man from the axis of rotation)

Substituting these values into the equation:

ω' = (43.0 kg * 0.16 rad/s + 0.5 * 5.99 × 10^2 kg * (3.87 m)^2 * 0.16 rad/s - 0.5 * 5.99 × 10^2 kg * (0.64 m)^2 * ω') / 43.0 kg

Solving for ω', we can simplify the equation to:

ω' = (43.0 kg * 0.16 rad/s + 0.5 * 5.99 × 10^2 kg * (3.87 m)^2 * 0.16 rad/s) / (43.0 kg + 0.5 * 5.99 × 10^2 kg * (0.64 m)^2)

Evaluating this expression, we get the value of the new angular speed ω'.