A playground merry-go-round has a disk-shaped platform that rotates with negligible friction about a vertical axis. The disk has a mass of 240 kg and a radius of 1.4 m. A 33- kg child rides at the center of the merry-go-round while a playmate sets it turning at 0.20rpm. If the child then walks along a radius to the outer edge of the disk, how fast will the disk be turning? (answer must be in rpm)

http://www.jiskha.com/display.cgi?id=1258578369

i tried the technique that was posted on there, but the answer was wrong. and my answer is supposed to be in RPM.

can somebody explain how to do this problem without changing it to radians? the final answer has to be in rates per minute

To solve this problem, we can use the principle of conservation of angular momentum. The initial angular momentum of the system is equal to the final angular momentum of the system.

Before the child moves to the outer edge, the initial angular momentum is given by:

L_initial = I_initial * ω_initial

where L_initial is the initial angular momentum, I_initial is the initial moment of inertia, and ω_initial is the initial angular velocity.

The final angular momentum can be calculated as:

L_final = I_final * ω_final

where L_final is the final angular momentum, I_final is the final moment of inertia, and ω_final is the final angular velocity.

Since there is negligible friction, the total angular momentum of the system remains constant. Therefore, we have:

L_initial = L_final

Now let's calculate the initial and final angular momenta:

The initial moment of inertia, I_initial, can be calculated using the formula for a solid disk:

I_initial = (1/2) * m * r_initial^2

where m is the mass of the disk and r_initial is the initial radius of the child when he is at the center of the merry-go-round.

Substituting the given values:

I_initial = (1/2) * 240 kg * (1.4 m)^2 = 235.2 kg * m^2

The final moment of inertia, I_final, can be calculated using the formula for a solid disk as well:

I_final = (1/2) * m * r_final^2

where r_final is the final radius of the child when he moves to the outer edge of the disk.

The moment of inertia of the disk remains the same as its mass is unchanged. Therefore, I_final is also equal to 235.2 kg * m^2.

Since ω_initial is given as 0.20 rpm, we need to convert it to radians per second (rad/s). There are 2π radians in one revolution, so:

ω_initial = (0.20 rpm) * (2π rad/1 min) * (1 min/60 s) = 0.0209 rad/s

Using the conservation of angular momentum equation, we can solve for ω_final:

L_initial = L_final
I_initial * ω_initial = I_final * ω_final

Substituting the values:

(235.2 kg * m^2) * (0.0209 rad/s) = (235.2 kg * m^2) * ω_final

Simplifying:

ω_final = (235.2 kg * m^2) * (0.0209 rad/s) / (235.2 kg * m^2)
ω_final = 0.0209 rad/s

Finally, we convert ω_final back to rpm:

ω_final = 0.0209 rad/s * (1 min/60 s) * (1 rev/2π rad) = 0.0033 rpm

Therefore, when the child walks to the outer edge of the disk, the merry-go-round will be turning at approximately 0.0033 rpm.