# Conservation of Rotational Momentum Merry-Go-Round Problem

• Freye
In summary, the conservation of angular momentum applies in both cases for the system of the merry-go-round and people. In case A, the people had zero angular momentum and added moment of inertia to the merry-go-round. On board, they gained angular momentum and took it away when jumping off radially. The total angular momentum of the system remains constant, and therefore the angular velocity of the merry-go-round changes to compensate for the loss of angular momentum. In case B, the people already have angular momentum when they step on the merry-go-round, and they preserve this momentum when they jump off radially, transferring it to the ground. Therefore, there is no change in the angular velocity of the merry-go-round. This demonstrates the conservation of angular
Freye
1. A 4.5m diameter merry-go-round is rotating freely with an angular velocity of 0.70 rad/s; its total moment of inertia is 1750 kg X m/s^2. Four people standing on the ground each of 65 kg mass suddenly step onto the edge of the merry-go-round.
a) What will be the angular velocity of the merry-go-round now? b)What if the people were on it initially, and then jumped off?

r= d/2 = 4.5/2 = 2.25m
w$$_{}1$$= 0.70 rad/s
I$$_{}1$$= 1750kg X m^2
I$$_{}people$$ = 65 X 4 X 2.25^2 = 1316.25kg X m^2
I$$_{}2$$= I$$_{}1$$ + I$$_{}people$$ = 3066.25 Kg X m^2

2. I$$_{}1$$w$$_{}1$$ = I$$_{}2$$w$$_{}2$$
T=rFsin$$\theta$$
T=I$$\alpha$$

3. The correct answers as they are given in the back of the book are a) w$$_{}2$$=0.40rad/s and b)no change (if jump off radially)

a)
w$$_{2}$$ = I$$_{}1$$w$$_{}1$$/I$$_{}2$$
= (1750)(.07)/(3066.25)

b)
w$$_{}1$$= 0.40 rad/s [from part a)]
I$$_{}1$$ and I$$_{}2$$ from part a) switch places
w$$_{}2$$=I$$_{}1$$w$$_{}1$$/I$$_{}2$$
w$$_{}2$$ = (3066.25)(0.40)/(1750)
w$$_{}2$$= 0.70 rad/s [which is to be expected, however apparently incorrect]

At this point, I realized that for some reason the conservation of angular momentum must not apply in this scenario, and so I did

$$\theta$$ = 0degrees (the angle between the radius and the applied force, assuming they jump of radially)
T=rfsin$$\theta$$=0
$$\alpha$$=T/I
$$\alpha$$=0
Therefore $$\Delta$$w=0 [the correct answer]

Essentially, my question is why the law of conservation of angular momentum applies when people step onto the merry-go-round, but not when they step off? I know I am making some sort of obvious logical blunder, but I cannot see it :(.

The conservation of angular momentum applies in both cases for the system marry-go-round and people. In case A, the people had zero angular momentum and added moment of inertia to the merry-go- round. On board, they gained angular momentum and took it away, when jumping off radially.

ehild

Ehild, I agree with what you are saying. But if I am correct, this would suggest that the Merry-Go-Round should speed back up once the people have jumped off. However, according to the answers in my textbook, this is not the case. And using the Torque equation obviously gives us an angular acceleration of 0 because no force is applied. I am utterly perplexed

No, the merry go round would not speed back. The people had angular momenta and preserved it when jumping off radially with respect of the rotating frame of reference. So they took away
Ipeople2=1316.25*0.4 =526.5 kg m2 s-1 from the merry-go-round. The angular momentum of the merry-go-round + people was 1225 kg m2 s-1 before the people jumped off, it of the merry-go-round became 1225-526.5=698.5 after the people left. Divided by the moment of inertia of the merry-go round, you get 0.4 s-1 for the angular velocity. The people preserved their angular momenta till they were flying in air. They jumped off radially in the rotating frame of reference, but they had tangential velocity components with respect to the ground. When they reached the ground they transferred their angular momenta to the Earth.

ehild

ehild said:
No, the merry go round would not speed back. The people had angular momenta and preserved it when jumping off radially with respect of the rotating frame of reference. So they took away
Ipeople2=1316.25*0.4 =526.5 kg m2 s-1 from the merry-go-round. The angular momentum of the merry-go-round + people was 1225 kg m2 s-1 before the people jumped off, it of the merry-go-round became 1225-526.5=698.5 after the people left. Divided by the moment of inertia of the merry-go round, you get 0.4 s-1 for the angular velocity. The people preserved their angular momenta till they were flying in air. They jumped off radially in the rotating frame of reference, but they had tangential velocity components with respect to the ground. When they reached the ground they transferred their angular momenta to the Earth.

ehild
I think that I understand what you are saying

Essentially for part b)

L1 = 1225 kg m^2/s
L2 = L1 -526.5 = 698.5 kg m^2/s
L2 = I_merrygoround * w2
w2 = L2/I_merrygoround
w2 = 698.5/1750 = 0.40 rad/s

So essentially, the Angular Momentum when the people are on the Merry Go Round does not stay the same when they jump off of it, making L1 not equal to L2. Why then in part a) when they jump on does L1 = L2, thus allowing the equation I1w1=I2w2 to apply?

You have to count the total angular momentum that of the merry-go-round and people: it is conserved. Initially, the people stood on the ground, with zero speed and angular momentum. So the total angular momentum was equal to that of the merry-go-round. The people stepped on the marry-go-round and shared its angular momentum. When jumping off, they preserved their angular momentum so the marry-go-round lost that amount, but the total angular momentum stayed the same.

ehild

Thank you very much ehild. I completely understand know, and it makes sense that that should be the case. Thanks so much again for your time and explanation. I just wish that my textbook could have explained that a little better in the first place

You are welcome.

ehild

## 1. What is the conservation of rotational momentum?

The conservation of rotational momentum is a fundamental law of physics that states that the total rotational momentum of a system remains constant unless acted upon by an external torque. This means that the momentum of a rotating object will not change unless an external force is applied.

## 2. How does the merry-go-round problem relate to conservation of rotational momentum?

The merry-go-round problem is a classic example used to demonstrate the conservation of rotational momentum. It involves a rotating platform with people standing at different distances from the center. When the people move closer or farther from the center, the rotational speed of the platform changes, but the total momentum remains constant.

## 3. What factors affect the conservation of rotational momentum in the merry-go-round problem?

The conservation of rotational momentum in the merry-go-round problem is affected by the mass of the objects on the platform, their distance from the center of rotation, and the initial rotational speed of the platform. Any changes in these factors will result in a change in the platform's rotational momentum.

## 4. How can the conservation of rotational momentum be applied in real-life situations?

The conservation of rotational momentum is applicable in various scenarios, such as sports, transportation, and engineering. In sports, athletes use this principle to perform tricks and stunts, such as spinning on ice skates or throwing a discus. In transportation, the conservation of rotational momentum is used to design vehicles with stable handling, such as bicycles and cars. In engineering, it is used to design machines with rotating parts, such as turbines and engines.

## 5. Is the conservation of rotational momentum always true?

Yes, the conservation of rotational momentum is a fundamental law of physics and is always true in a closed system. However, it may not hold true in open systems where external forces, such as friction, air resistance, or torque, are present. In such cases, the total rotational momentum may change due to the external forces acting on the system.

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