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?
conservation of momentum applies, in this case, angular momentum.
Diskmomentum=I*w
Boy momentum=33*r*2PI*.2/60
set them equal, w will be in rad/sec
disk moment of inertial is 1/2*240*r if you didn't know. Thin disk.
the answer has to be in rpm, so does that mean that i'm not supposed to multiply by 2PI/60?
To find the angular velocity of the disk after the child moves to the outer edge, we can apply the law of conservation of angular momentum.
The angular momentum of an object is given by the product of its moment of inertia and angular velocity. In this case, the moment of inertia of the disk can be calculated using the formula for a solid disk:
I = (1/2) * m * r^2
where I is the moment of inertia, m is the mass, and r is the radius. Substituting the given values:
I = (1/2) * 240 kg * (1.4 m)^2
= 235.2 kg*m^2
Initially, the child is at the center of the merry-go-round, so the moment of inertia of the system is just that of the disk:
I_initial = 235.2 kg*m^2
When the child moves to the outer edge of the disk, the moment of inertia of the system changes. The moment of inertia now includes both the disk and the child. Using the same formula:
I_final = (1/2) * (240 kg + 33 kg) * (1.4 m)^2
= 242.22 kg*m^2
According to the conservation of angular momentum, the initial angular momentum is equal to the final angular momentum:
I_initial * ω_initial = I_final * ω_final
Solving for ω_final (the final angular velocity):
ω_final = (I_initial * ω_initial) / I_final
Substituting the given values:
ω_final = (235.2 kg*m^2 * (0.20 rpm * 2π rad/min)) / 242.22 kg*m^2
To simplify the calculations, we need to convert the angular velocity from rpm to rad/min:
ω_final = (235.2 kg*m^2 * (0.20 * 2π rad/min)) / 242.22 kg*m^2
Simplifying further, we find:
ω_final ≈ 1.287 rad/min
Therefore, the disk will be turning at approximately 1.287 rad/min when the child walks to the outer edge.