Conservation of Rotational Momentum Merry-Go-Round Problem

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  • #1
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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[tex]_{}1[/tex]= 0.70 rad/s
I[tex]_{}1[/tex]= 1750kg X m^2
I[tex]_{}people[/tex] = 65 X 4 X 2.25^2 = 1316.25kg X m^2
I[tex]_{}2[/tex]= I[tex]_{}1[/tex] + I[tex]_{}people[/tex] = 3066.25 Kg X m^2



2. I[tex]_{}1[/tex]w[tex]_{}1[/tex] = I[tex]_{}2[/tex]w[tex]_{}2[/tex]
T=rFsin[tex]\theta[/tex]
T=I[tex]\alpha[/tex]



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

a)
w[tex]_{2}[/tex] = I[tex]_{}1[/tex]w[tex]_{}1[/tex]/I[tex]_{}2[/tex]
= (1750)(.07)/(3066.25)
= 0.40 rad/s [the correct answer]

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

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

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

Answers and Replies

  • #2
ehild
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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
 
  • #3
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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
 
  • #4
ehild
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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
 
  • #5
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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?
 
  • #6
ehild
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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
 
  • #7
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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
 
  • #8
ehild
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You are welcome.:smile:

ehild
 

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