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 230 kg and a radius of 1.6 m. A 38- kg child rides at the center of the merry-go-round while a playmate sets it turning at 0.15rpm. If the child then walks along a radius to the outer edge of the disk, how fast will the disk be turning (in rpm)?
i am aware that another question has been posted that is exactly the same as this, but when i used that technique, it did not work for me, so i am posting this question again. like i said, i need to get this answer in RPM and NOT rad/s
Convert rpm to rad/s to do the problem using conservation of angular momentum. Covert back to rpm when you get the answer. The child increases the moment of inertia by walking to the outer edge of the disk. I assume you know the formula for the moment of inertia of the disk
i tried that, but i'm not getting the answer right
i did disk momentum=Iw
and boy momentum=m*rpm*2pi/60
then i set those two equal to each other and then changed the w back to rpm, but i'm not getting the right answer.
To solve this problem, we can make use of the principle of conservation of angular momentum. According to the principle, the initial angular momentum of the system must be equal to the final angular momentum.
The equation for angular momentum is given by:
L = Iω
Where:
L = Angular momentum
I = Moment of inertia
ω = Angular velocity
Initially, with only the child on the merry-go-round, the moment of inertia of the system is given by:
I_initial = MR²
Where:
M = Mass of the disk
R = Radius of the disk
Given the mass of the disk, M = 230 kg, and the radius of the disk, R = 1.6 m, we can calculate the initial moment of inertia:
I_initial = 230 kg * (1.6 m)² = 588.8 kg·m²
Now, let's consider the final situation when the child walks to the outer edge of the disk. The total moment of inertia of the system can be calculated as the sum of the moment of inertia of the disk and the moment of inertia of the child.
I_final = (M + m) * R²
Where:
M = Mass of the disk
m = Mass of the child
R = Radius of the disk
Given the mass of the disk, M = 230 kg, the mass of the child, m = 38 kg, and the radius of the disk, R = 1.6 m, we can calculate the final moment of inertia:
I_final = (230 kg + 38 kg) * (1.6 m)² = 334.56 kg·m²
Since angular momentum is conserved, we can equate the initial and final angular momenta:
I_initial * ω_initial = I_final * ω_final
Initially, the angular velocity is given as 0.15 rpm. But we need to convert it to rad/s to use it in the equation. Since 1 revolution = 2π radians, we have:
ω_initial = (0.15 rev/min) * (2π rad/1 rev) * (1 min/60 s) = 0.0157 rad/s
Now, we can rearrange the equation to solve for the final angular velocity, ω_final:
ω_final = (I_initial * ω_initial) / I_final
Substituting the values into the equation:
ω_final = (588.8 kg·m² * 0.0157 rad/s) / 334.56 kg·m²
≈ 0.0277 rad/s
Finally, to convert the angular velocity from rad/s to rpm, we can use the conversion factor:
1 rad/s = (1 rev/2π rad) * (60 s/1 min)
ω_final = 0.0277 rad/s * (1 rev/2π rad) * (60 s/1 min)
≈ 0.0832 rpm
Therefore, the final angular velocity of the merry-go-round when the child walks to the outer edge of the disk is approximately 0.0832 rpm.
To find the final rotational speed of the disk in rpm, we can use the principle of conservation of angular momentum. The angular momentum of the system before and after the child walks to the outer edge of the disk should be equal.
The angular momentum (L) of an object is given by the formula:
L = I * ω,
where I is the moment of inertia and ω is the angular velocity.
Before the child walks to the outer edge, the moment of inertia of the system is calculated by considering the child at the center of the merry-go-round:
I1 = m1 * r1^2,
where m1 is the mass of the child and r1 is the initial radius.
Now, let's calculate the initial angular momentum:
L1 = I1 * ω1,
where ω1 is the initial angular velocity given in the problem (0.15 rpm).
After the child walks to the outer edge, the moment of inertia of the system changes because the mass is distributed at a larger radius:
I2 = (m1 + m2) * r2^2,
where m2 is the mass of the disk and r2 is the final radius (which is the sum of the initial radius and the radius to the outer edge).
Since the angular momentum is conserved, we can set L1 equal to L2 and solve for the final angular velocity (ω2):
L1 = L2
I1 * ω1 = I2 * ω2
m1 * r1^2 * ω1 = (m1 + m2) * r2^2 * ω2
Given the values from the problem, substitute the values:
(38 kg) * (0.8 m)^2 * (0.15 rpm) = (38 kg + 230 kg) * (2.4 m)^2 * ω2
Now, solve for ω2:
ω2 = [(38 kg) * (0.8 m)^2 * (0.15 rpm)] / [(38 kg + 230 kg) * (2.4 m)^2]
Calculate the value of ω2 and express it in rad/s:
ω2 ≈ 1.02 rad/s
Finally, convert the angular velocity from rad/s to rpm by dividing by (2π radians) and multiplying by (60 seconds):
Final angular velocity in rpm ≈ (1.02 rad/s) * (60 s) / (2π radians) ≈ 9.70 rpm
Therefore, the final rotational speed of the disk, when the child walks to the outer edge, is approximately 9.70 rpm.