Solve for wInertia Confusion: Calculating Speed of a Merry-Go-Round

In summary, Andre found that when the kid is on the outside of the merry-go-round, the merry go round will be turning faster, and when the kid is in the middle of the merry-go-round, the merry go round will be turning slower. However, he is still confused about inertia.
  • #1
tucky
30
0
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
 
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  • #2
Seems OK sofar. Now what is I of the total system: disk + kid on the outside and the total I of the disk + kid in the centre. If the kid is sitting in the centre then obviously there is much 'r' left. :smile:
 
  • #3
Andre,

I am sorry I still don’t understand. Did you mean that when the kid is in the center that you do not calculate radius in the inertia equation (I=½mr^2), or do you not square the radius? Thank you for your quick response!
 
  • #4
No, and I take back the "seems to be alright". let's focus on the kid first. What would be his I, when sitting on the edge of the disk? Think of a mass concentrated in one point. And what would be his turning energy over there when the merry go round goes round?

And incidentally perhaps recheck your formulas. What is the definition of I? the turning momentum and what is the formula for energy of a turning mass?
 
  • #5
Inertia with kid on side = 45.39kg (1.52m)^2 =104.8kgm^2
Inertia with kid in the middle = 158.9 (1.52m)^2 = 367kgm^2

Is that right?

So, you think it would be better to use the momentum equation L = I*omega or to use the KE=½*I*omega^2

I am sorry about my confusion.
 

FAQ: Solve for wInertia Confusion: Calculating Speed of a Merry-Go-Round

What is inertia?

Inertia is the resistance of an object to change its state of motion. In simpler terms, it is the tendency of an object to stay at rest or continue moving in a straight line at a constant speed.

What is the formula for calculating inertia?

The formula for calculating inertia is I = mr², where m is the mass of the object and r is the distance of the object from the axis of rotation. This formula is used to calculate the moment of inertia, which is a measure of an object's resistance to rotational motion.

What is the speed of a merry-go-round?

The speed of a merry-go-round can vary depending on its size and the force applied to it. However, a typical speed for a merry-go-round is around 10-15 miles per hour.

How do you calculate the speed of a merry-go-round?

To calculate the speed of a merry-go-round, you need to know the radius of the merry-go-round and the time it takes for one full rotation. The formula for calculating speed is v = 2πr/t, where v is the speed, r is the radius, and t is the time.

How does inertia affect the speed of a merry-go-round?

Inertia plays a crucial role in the speed of a merry-go-round. The greater the inertia of the merry-go-round, the more force is required to change its speed. This means that a merry-go-round with a larger mass or a larger distance from the axis of rotation will have a slower speed compared to one with a smaller mass or distance.

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