- #1
tucky
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Thanks to gnome for answering my last physics question. However, I am still confused about inertia. Here is my next question maybe someone can alleviate my confusion:
Q: A merry-go-round (a piece of playground equipment)consists of a disk 10 ft in diameter that weighs 250 lbs. The disk turns on a low-friction bearing.
A kid who weighs 100 lbs sits on the edge of the merry-go-round. The merry-go-round is turning at 1 rev every 3 seconds. If the kid moves to the center of the merry-go-round, how fast will the merry-go-round be turning?
A: I know that when the kid moves to the center the merry-go round is going to move at a faster speed. I am not sure how to figure it mathematically. I am confused how to calculate the inertia when the kid is on the outer part of the merry-go-round and when the kid moves toward the center.
5ft = 1.52ft; 250lb = 113.5kg; 100lb = 45.39kg; w = .667rad/s
I think the inertia of a disk is I =½ m r^2, I do not know how to account for the kid when is on the outside and when he is on the inside.
Then I was going to set the KE equations when the kid is on the outside equal when the kid is in the middle to find the speed.
KE(kid on outside) = KE(kid on inside)
½ I(kid on outside) (.667rad/s)^2 = ½ I(kid on inside) w^2
Q: A merry-go-round (a piece of playground equipment)consists of a disk 10 ft in diameter that weighs 250 lbs. The disk turns on a low-friction bearing.
A kid who weighs 100 lbs sits on the edge of the merry-go-round. The merry-go-round is turning at 1 rev every 3 seconds. If the kid moves to the center of the merry-go-round, how fast will the merry-go-round be turning?
A: I know that when the kid moves to the center the merry-go round is going to move at a faster speed. I am not sure how to figure it mathematically. I am confused how to calculate the inertia when the kid is on the outer part of the merry-go-round and when the kid moves toward the center.
5ft = 1.52ft; 250lb = 113.5kg; 100lb = 45.39kg; w = .667rad/s
I think the inertia of a disk is I =½ m r^2, I do not know how to account for the kid when is on the outside and when he is on the inside.
Then I was going to set the KE equations when the kid is on the outside equal when the kid is in the middle to find the speed.
KE(kid on outside) = KE(kid on inside)
½ I(kid on outside) (.667rad/s)^2 = ½ I(kid on inside) w^2