Conservation of Angular Momentum -- Child jumping onto a Merry-Go-Round

[Mustard]

Homework Statement

A merry-go-round of radius R = 2.0 m has a moment of inertia I = 250 kg⋅m^2 and is rotating at 10 rpm (revolutions per minute). A child whose mass is 25 kg jumps onto the edge of the merry-go-round, heading directly toward the center at 6.0 m/s. What is the new angular
speed (in rpm) of the merry-go-round?

Relevant Equations

I=mr^2
v=wr

So we know that the initial intertia of the merry go round is 250 kg m^2 and its angular speed is 10 rpm. MGRs angular momentum would be L=Iw=250(10)=2500kg m^2 rpm.

We know the mass if the child is 25kg, and the child's linear velocity is 6m/s. We convert linear to angular w= v/r = 6/2 = 3rad/s=28.62rpm. Inerta of child = mr^2 =25(2)^2=100kg m^2. So the child angular momentum is L=Iw=100(28.62)=2862kg m^2 rpm.

L1 + L2 = L3
L1 + L2 = (Inertia1 + Inertia2)w
2500+2862=(250+100)w
w=5362/350=15.32 rpm.

Am I doing this right?

 

[haruspex]

the child's linear velocity is 6m/s. We convert linear to angular w= v/r = 6/2 = 3rad/s=28.62rpm

Angular velocity is always about some axis. How do you convert linear velocity to angular velocity about a specified axis?

 

[Frodo]

Do the problem in small steps.

Step 1. What happens after the child steps on the roundabout but before she has left the rim?

Step 2. What happens as she moves from the rim towards the centre?

Step 3. What happens when the child reaches the centre? This bit is more tricky as you need to assume what "reaching the centre" means. The child has a physical size so you need to assume a moment of inertial value for the child; or you assume she has zero dimension but she retains her angular momentum.

 

[jbriggs444]

The problem does not involve the child moving to the center of the carousel.

The specification "directly toward the center" is there to establish the direction the child jumps. The child lands on the rim of the carousel in an inelastic collision and moves with the rim of the carousel as a result of the collision.

[This question was discussed to death recently with pretty much everyone latching onto the unintended interpretation that the child continues walking forward]

 

[Frodo]

I am confused.

A child whose mass is 25 kg jumps onto the edge of the merry-go-round, heading directly toward the center at 6.0 m/s

If the jump is radial the velocity is not necessary - the child lands on the rim without adding angular momentum due to her velocity.. So why give the radial velocity of 6.0 m/s unless the child is intended to continue?

 

[jbriggs444]

I am confused.If the jump is radial the velocity is not necessary - the child lands on the rim. So why give the radial velocity of 6.0 m/s unless the child is intended to continue?

Confusing extraneous information! Always a good policy to weed out the folks who do not really understand what they are doing.

If the child ends at the center, the initial 6.0 m/s is also irrelevant.

 

[Frodo]

A child ... jumps onto the edge of the merry-go-round, heading directly toward the center at 6.0 m/s

I suggest that the word heading invites the interpretation of continued motion as in "The man is heading for the train station at 6 m/s".

It's a badly worded question.

 

[etotheipi]

As has been mentioned above, the only reason why the question contains the wording "heading directly toward the centre" is to emphasise that the person jumping has no initial angular momentum about the central axis of the merry-go-round, which would otherwise need to be accounted for. It specifically says that he jumps onto the edge.

 

[jbriggs444]

"Heading" borrows a bit from the nautical -- a direction rather than a destination. [heading, bearing and course are all nicely differentiated in nautical usage].

"Heading toward" carries less of an implication of continuation than "heading to" in my mind.

Jumping "onto the edge" carries a pretty strong implication against continued motion.

But, as general advice: State your assumptions up front and then answer the clarified problem. That is the safe approach.

 

[Frodo]

Always a good policy to weed out the folks who do not really understand what they are doing.

May I suggest that you think a little more about the angular momentum the child has when she lands on the rim.

Travelling at 6 metres/sec she possesses a huge amount of angular momentum about her feet and will therefore either

a) rotate about her feet and fall towards the centre, in which case her centre of mass is not on the rim; or
b) step towards the centre in order to prevent her falling, in which case her centre of mass is not on the rim.

In both cases the presumed intent of having the child's mass remain on the rim is incorrect.

It's a very badly worded question.

 

[jbriggs444]

May I suggest that you think a little more about the angular momentum the child has when she lands on the rim.

I do not think we are expected to get into 3 dimensions here. But if you insist, gravity cancels that unaligned angular momentum nicely. The child naturally leans backward as we all do when landing a jump... Or grabs a stanchion. Or tumbles to rest with the forward roll canceled and the angular momentum absorbed by the axle which is naturally unable to tolerate rotation in the direction of that forward roll.

Been there, done a similar stupid pet trick -- the stanchions hurt.

Standing in the middle of a slightly inclined playground carousel in Bettendorf, Iowa, walking uphill as if on a treadmill. The thing gets spinning pretty fast and I finally start contemplating exit strategies. I could switch to the other side of the axle and try slowing the thing down. But then I'd be walking backward and I'm already pretty close to losing balance. I could simply drop to the ground. But I'd be picked up by the carousel surface and the effect would be essentially identical to losing balance. I could call for help. But it wouldn't arrive in time and I'd look stupid. I could shift closer to the axle and try to buy time to think. I knew that if I lost balance I must try to induce as little horizontal motion as possible. The decision made itself in pretty short order. Balance lost and ribs painfully bruised. Lesson learned.

 

[haruspex]

May I suggest that you think a little more about the angular momentum the child has when she lands on the rim.

Travelling at 6 metres/sec she possesses a huge amount of angular momentum about her feet and will therefore either

a) rotate about her feet and fall towards the centre, in which case her centre of mass is not on the rim; or
b) step towards the centre in order to prevent her falling, in which case her centre of mass is not on the rim.

In both cases the presumed intent of having the child's mass remain on the rim is incorrect.

It's a very badly worded question.

i don't see a problem angular momentum about the feet. Merry-go-rounds are not flat platforms; they have handholds. Indeed, I see no issue with the wording, it is quite clear.
But linear momentum is a worry. Jumping on at over 20km/h will end in tears.

 

[hmmm27]

A merry-go-round of radius R = 2.0 m has a moment of inertia I = 250 kg⋅m^2 and is rotating at 10 rpm (revolutions per minute). A child whose mass is 25 kg jumps onto the edge of the merry-go-round, heading directly toward the center at 6.0 m/s. What is the new angular speed (in rpm) of the merry-go-round?

Doesn't make any sense.

If you assume the child is a stick-figure line, then - once the child reaches the center - the answer doesn't require any calculation. Otherwise you need to know the width of the child (which is probably politically incorrect);

If you assume the child just jumped onto the edge and stayed there, then there's no point to "6m/s".

Either way, that kid had waaay too much caffeine.

 

[jbriggs444]

If you assume the child just jumped onto the edge and stayed there, then there's no point to "6m/s".

And the point of the 6 m/s if the child makes it to the center is...?

 

[hmmm27]

And the point of the 6 m/s if the child makes it to the center is...?

How far away you can hear the thud as the child hits the pole ?

 

[jbriggs444]

How far away you can hear the thud as the child hits the pole ?

Typically there is no pole. Flat plate in the middle. The bearings are on the under-side. You do not see them much in U.S. playgrounds these days. Perhaps a bit too dangerous for the modern risk-averse world. But my, oh my, the ones with good bearings were fun!

 

[hutchphd]

Standing in the middle of a slightly inclined playground carousel in Bettendorf, Iowa, walking uphill as if on a treadmill.

I have a slightly weird question. I, too. recall them happily from childhood but I was in high school before I realized the truly suicidal method of pumping them like a swing. After the initial push one can hang out at the edge of the inclined disc extending out at the high point of the revolution and hauling your butt inboard at the bottom. In no time you can get going so fast you can no longer make the inward haul. I never discovered this method as a kid...probably a good thing. Did you?

 

[jbriggs444]

I have a slightly weird question. I, too. recall them happily from childhood but I was in high school before I realized the truly suicidal method of pumping them like a swing. After the initial push one can hang out at the edge of the inclined disc extending out at the high point of the revolution and hauling your butt inboard at the bottom. In no time you can get going so fast you can no longer make the inward haul. I never discovered this method as a kid...probably a good thing. Did you?

Never did. Though we had a back-yard swing with long chains hung from a strong tree with a wooden seat you could stand on. My dad had put 5/8 inch eye bolts through the limb where the chains attached so that the eyes were level with each other. Absolutely the best swing I have ever used. It might still be in place now, 60 years later, were it not for Dutch Elm disease having taken that tree. Pumping that from a standing posture at high amplitude involved rising from a crouch to an upright posture at the bottom of the arc. Being fifteen feet in the air, slightly higher than the tree limb with slack chains was a cool feeling.

Then there was the time the fifth graders from my dad's school came to visit. He had to put the climbing rope (same tree, much higher limb) away because the stunts that my brother and sister and I (1st, 3rd and 4th graders) were doing were too risky for the visitors to attempt.

 

[Frodo]

Doesn't make any sense.

I totally agree.

Either way, that kid had waaay too much caffeine.

She needed the all the caffeine she could get because, if she stopped on the rim she would have needed to decelerate in, say, 0.3m.

She would have pulled about 6g by applying a force of about 150kg.

Not bad for a 7 year old (the 50 percentile for 25kg girls is 7 1/2 years).

And there's me thinking little girls are made of "sugar and spice and all things nice". I didn't realize they were built like tanks.