A 4.2-m-diameter merry-go-round is rotating freely with an angular velocity of 0.80 rad/s. Its total moment of inertia is . Four people standing on the ground, each of mass 65 kg, suddenly step onto the edge of the merry-go-round.
- What is the angular velocity of the merry-go-round now?
- What if the people were on it initially and then jumped off in a radial direction (relative to the merry-go-round)?
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This is Giancoli Answers with Mr. Dychko. The angular momentum is conserved in this question because when the people step onto the edge of the merry-go-round, they don't exert any torques in any direction they are just, it's basically like a collision just an inelastic collision where they just stick to the merry-go-round and that increases the merry-go-round's moment of inertia and so it will slow it down but the angular momentum is conserved. So angular momentum is moment of inertia final times final angular velocity and that equals moment of inertia initially which is just that of the merry-go-round times its initial angular velocity. So solve for ω f by dividing both sides by I m plus 4 times mass of a person times the radius of the merry-go-round squared; each person can be considered a point mass and so the moment of inertia of each person is their mass times radius squared and it's four people so we multiply that by 4. So we have final angular velocity is 1360 kilogram meter squared—moment of inertia of the merry-go-round— times 0.8 radians per second— initial angular velocity— divided by 1360 kilogram meter squared plus 4 times 65 kilograms times 4.2 meter—diameter—divided by 2 to get the radius and then square that result and that's 0.43 radians per second. In part (b), it's the same idea that the angular momenta are the same. So angular momentum is conserved because when they step off the merry-go-round, they are not exerting any torque, they step off radially and so there's no component of their force perpendicular to the radius. Does that make sense? So they are not making the merry-go-round spin in any sense either faster or slower when they step off radially. So that means angular momentum is still conserved because there's no net torque in other words. So the moment of inertia final times angular velocity final equals moment of inertia initial times angular velocity initial. So initially, they are starting on the merry-go-round so the total moment of inertia is that of the merry-go-round alone plus 4 times the moment of inertia of each of these point masses: mass of the person times radius of the merry-go-round squared times initial angular velocity and then after they step off, the total moment of inertia is just that of the merry-go-round. So we'll solve ω f by dividing both sides by I M and so we have 1360 kilogram meter squared plus 4 times 65 times 4.2 over 2 squared times 0.8 radians per second divided by 1360 kilogram meter squared and that gives 1.5 radians per second. And so as expected, the merry-go-round will increase its angular velocity since it has decreased its moment of inertia.
Dear Mr. Dychko. First of all, I would like to express my uttermost gratitude for all your help - your videos have significantly helped me towards comprehending some important criteria towards certain problems - and I hope you, as well as your family, are doing well during this pandemic. In the second part of the problem, when I submitted my answer on Pearson, it told me that the answer for the final angular velocity is equal to 0.80 rad/s, thus the angular velocity does not change when the people jumped off the merry-go-round. Was there an element that you missed in your video, and if this is the case, please let me know. Once again, thank you for your help, and as always, have a wonderful day. Stay safe.
Thanks for that question. I've been beating my head against the wall. The angular velocity does NOT change, or at least this particular problem doesn't think it does.
Mr. Dychko. I think your website is great. Your explanations have been very helpful. Here is my thought on why the final angular velocity is equal to 0.80 rad/s:
One might reason that since no net torque acts on the merry-go-round, that the system's total moment of inertia decreases as the four people jumped off radially, and expecting that its angular speed of the merry-go-round should increase.
However, the angular momentum of the system (the merry-go-round and the four people jumping off) is conserved. As the four people jump off, they each carry angular momentum with them. If you consider the merry-go-round and the four people as two separate systems, each with angular momentum from their moments of inertia and angular speed, it is easy to see that by dropping the mass of the four people off the merry-go-round, no net external torque acts on the merry-go-round and so therefore its moment of inertia does not change, and therefore its angular speed will not change.
Note that the angular momentum of the four people also does not change until they hit the ground and at that time, friction (external torque) stops their motion.
This situation is just different than (rather than being the opposite of part (a)) the four people jumping on the merry-go-round because they jumped on radially with no angular momentum or velocity. So the merry-go-round’s moment of inertia increased, but its velocity decreased as the people jumped on with no angular momentum or velocity.
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