Angular Momentum Question concerning a Merry Go Round

[WilkinsD]

Homework Statement

If the steel disk has mass of 200 kg and a radius of 2 meters you can make it spin by applying a force to the rim. This torque increases the angular momentum of the disk. Suppose the force is 20 Newtons. How long would you have to apply it to get the wheel spinning 5 times a minute?

What would happen to the rate of spin if you then jumped on the rim of the wheel with your mass of 60 kg?

Homework Equations

I=1/2mr2

I1*w0/(i1+i2)

The Attempt at a Solution

[/B]I think I have all the variables figured out I am just having an issue solving the questions.

I have the w=0.1rad/s , I =400 kg*m^2, t=40 N*m.
For the second problem I got the w2=0.07rad/s which would mean the merry go rounds rate of spin would decrease by .03 rad/s when the mass of 60kg jumps on.

(400)(0.1)/(320+240) = 40 / 560 =0.07rad/s
Any help would be appreciated thanks!

 

[TomHart]

I have the w=0.1rad/s

What does this represent? Hopefully not 5 revs/minute.

 

[WilkinsD]

What does this represent? Hopefully not 5 revs/minute.

No I believe that is my angular momentum before the 60kg person jumps on.

 

[TomHart]

No I believe that is my angular momentum before the 60kg person jumps on.

It has units of angular velocity though.

 

[WilkinsD]

It has units of angular velocity though.

Maybe that is where I have messed up. Most of the problems my teacher provided were not setup in this manner.

 

[TomHart]

You probably want to convert the angular velocity of 5 rev/minute to rad/s.

 

[WilkinsD]

It has units of angular velocity though.

The first time I solved it I got these as my answers but wasn't for sure if they were correct.

1. 300s

2. The acceleration would then be reduced to 0.0625 rad/s^2.

 

[WilkinsD]

You probably want to convert the angular velocity of 5 rev/minute to rad/s.

So 5 * 2pi? or 31.42 radians ?

 

[TomHart]

It's 31.42 radians/minute. Convert that to rad/s.

 

[WilkinsD]

It's 31.42 radians/minute. Convert that to rad/s.

Alright so that gives me .52 rad/s

 

[TomHart]

Yes, so then work with that to calculate the time it takes to go from ω=0 to ω=0.52 rad/s. That will be based on the angular acceleration you calculated from your torque, inertia, etc.

 

[WilkinsD]

Yes, so then work with that to calculate the time it takes to go from ω=0 to ω=0.52 rad/s. That will be based on the angular acceleration you calculated from your torque, inertia, etc.

I must be slow today lol
How long would you have to apply it to get the wheel spinning 5 times a minute? I can't wrap my head around the equation I need to solve to get how long I would have to apply the force to get it spinning 5 times a minute.

 

[TomHart]

What equations do you have to work with? If you remember the linear equations, it is very easy to figure out the angular motion equations.
Just substitute θ for x, ω for v, α for a, Γ for F, I for m. I think that covers it all.

 

[WilkinsD]

What equations do you have to work with? If you remember the linear equations, it is very easy to figure out the angular motion equations.
Just substitute θ for x, ω for v, α for a, Γ for F, I for m. I think that covers it all.

Yeah but what formula solves for how long to apply the force to get the wheel to turn 5 times a minute ?

 

[TomHart]

Yeah but what formula solves for how long to apply the force to get the wheel to turn 5 times a minute ?

I didn't memorize any equation that does that. But I know how to get angular acceleration from torque and moment of inertia.

 

[WilkinsD]

I didn't memorize any equation that does that. But I know how to get angular acceleration from torque and moment of inertia.

So would that just be my torque which is 40 N * m and my I which is 400 kg*m^2

40=400a so that gives me my a=0.1 rad/s ?

 

[TomHart]

Yes, but to be more accurate:
α = 0.1 rad/s^2

 

[PhysicsA1]

in which equation used uses α?

 

[WilkinsD]

5 times per minute would be 12 seconds per turn. so it should be .083/sec.

 

[WilkinsD]

in which equation used uses α?

Torque=I*a

 

[PhysicsA1]

5 times per minute would be 12 seconds per turn. so it should be .083/sec.

I figured that out exactly the same. So once we figure this out would be multiply the .083 through to find our final answer

 

[TomHart]

5 times per minute would be 12 seconds per turn. so it should be .083/sec.

You calculated number of revolutions per second, but you need radians/second.

 

[WilkinsD]

You calculated number of revolutions per second, but you need radians/second.

So that just brings me back to the 0.522 radians per second lol

 

[TomHart]

So that just brings me back to the 0.522 radians per second lol

Yes. That is the final angular velocity. So you know three things: initial angular velocity (0 rad/s), the final angular velocity (0.52 rad/s) and the angular acceleration (0.1 rad/s^2). From that you can calculate the time. So find an equation that has those three variables and time and you should be able to solve that equation for time.

 

[WilkinsD]

Yes. That is the final angular velocity. So you know three things: initial angular velocity (0 rad/s), the final angular velocity (0.52 rad/s) and the angular acceleration (0.1 rad/s^2). From that you can calculate the time. So find an equation that has those three variables and time and you should be able to solve that equation for time.

so I found

this was for sure not provided by my teacher. That would be 0.1 = .52/t which would make t = 5.2 seconds correct?

 

[TomHart]

I think that's right.

 

[WilkinsD]

I think that's right.

Oops had 5.25 seconds should be 5.2 seconds

 

[PhysicsA1]

Oops had 5.25 seconds should be 5.2 seconds

So with the addition of the 60kg would that then make the time be 6.75 rad/s instead of 5.2 rad/sec, which would make the rate of spin be slower!?
right?

 

[PhysicsA1]

I think that's right.

So with the addition of the 60kg would that then make the time be 6.75 rad/s instead of 5.2 rad/sec, which would make the rate of spin be slower!?
right?

 

[TomHart]

6.75 rad/s is a faster rotation than 5.2 rad/s. Since angular momentum has to be conserved, I would expect that increasing the moment of inertia would decrease the angular velocity. You should show your work to make it easier for people to help.

One thing that will really pay off in the long run in working physics problems is to be very deliberate in writing out complete, accurate equations and showing your step-by-step work process.

You know, I would do well to heed the above advice myself. It is good advice.

 

[PhysicsA1]

6.75 rad/s is a faster rotation than 5.2 rad/s. Since angular momentum has to be conserved, I would expect that increasing the moment of inertia would decrease the angular velocity. You should show your work to make it easier for people to help.

One thing that will really pay off in the long run in working physics problems is to be very deliberate in writing out complete, accurate equations and showing your step-by-step work process.

You know, I would do well to heed the above advice myself. It is good advice.

I= 520 kg*m (Due to taking the initial mass (200) and adding 60 to it due to us jumping on the merry go round) so...I=(1/2)(260)(2)^2
40=520a SO a=.077
SO .077=.55/t; Therefore T=6.75 sec.

Where did i go wrong? or misunderstand??

 

[haruspex]

I= 520 kg*m (Due to taking the initial mass (200) and adding 60

Do you have the shape of a flat disk?

40=520a

I do not understand this step. What calculation is this?
What law are you applying?

 

[PhysicsA1]

Are you the shape of a flat disk?

yes. its a playground merry go round

 

[haruspex]

yes. its a playground merry go round

That's the merry-go-round, not you, the 60kg passenger.

 

[PhysicsA1]

That's the merry-go-round, not you, the 60kg passenger.

Here are the questions again.

If the steel disk has mass of 200 kg and a radius of 2 meters you can make it spin by applying a force to the rim. This torque increases the angular momentum of the disk. Suppose the force is 20 Newtons. How long would you have to apply it to get the wheel spinning 5 times a minute?

What would happen to the rate of spin if you then jumped on the rim of the wheel with your mass of 60 kg?

 

[haruspex]

Here are the questions again.

If the steel disk has mass of 200 kg and a radius of 2 meters you can make it spin by applying a force to the rim. This torque increases the angular momentum of the disk. Suppose the force is 20 Newtons. How long would you have to apply it to get the wheel spinning 5 times a minute?

What would happen to the rate of spin if you then jumped on the rim of the wheel with your mass of 60 kg?

Yes, I read and understood the question.
When you calculated the modified moment of inertia for the last part, you treated it as though, in jumping on the disk, your 60kg became evenly spread over its surface. Not ideal behaviour for playground equipment.

 

[PhysicsA1]

Yes, I read and understood the question.
When you calculated the modified moment of inertia for the last part, you treated it as though, in jumping on the disk, your 60kg became evenly spread over its surface. Not ideal behaviour for playground equipment.

So it will only be effecting the side of which i jump on. How in return will that actually be calculated then?

 

[haruspex]

So it will only be effecting the side of which i jump on. How in return will that actually be calculated then?

The axis is still the post on which the merry-go-round spins. You have to add your moment of inertia about that axis to that of the disk.
What is the moment of inertia of a point mass rotating around an axis?

 

[PhysicsA1]

The axis is still the post on which the merry-go-round spins. You have to add your moment of inertia about that axis to that of the disk.
What is the moment of inertia of a point mass rotating around an axis?

I honestly have no clue

 

[haruspex]

I honestly have no clue

See http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html

 

[Caio Graco]

M = I.α => M = I.(Δω/Δt)=> Δt = I.Δω/M => Δt = 400 . (π/6) / 40 => Δt = 5π/3 s

I do not know if that's the way it is. What do you think?

 

[haruspex]

M = I.α => M = I.(Δω/Δt)=> Δt = I.Δω/M => Δt = 400 . (π/6) / 40 => Δt = 5π/3 s

I do not know if that's the way it is. What do you think?

You do not need any calculus. You already had that I₀ω₀=I₁ω₁, conservation of angular momentum. You were going wrong in calculating I₁, the sum of the disk's moment of inertia (I₀) and your moment of inertia about the disk's axis.
We can treat you as a point mass. What (I ask again) is the moment of inertia of a point mass m about an axis distance r from the mass? I gave you a link to look this up. What did you learn?

 

[Caio Graco]

You do not need any calculus. You already had that I₀ω₀=I₁ω₁, conservation of angular momentum. You were going wrong in calculating I₁, the sum of the disk's moment of inertia (I₀) and your moment of inertia about the disk's axis.
We can treat you as a point mass. What (I ask again) is the moment of inertia of a point mass m about an axis distance r from the mass? I gave you a link to look this up. What did you learn?

The conservation of angular momentum can only be used for the second question. In the first (for which I did the math) can not be conservation of the moment, as there is the actuation of an external torque.

 

[haruspex]

The conservation of angular momentum can only be used for the second question. In the first (for which I did the math) can not be conservation of the moment, as there is the actuation of an external torque.

I thought the first part was settled at post #25. I only came in at post #32, where the discussion had moved to the second question, so I assumed your post #41 was in relation to the second question.

 

[Caio Graco]

I thought the first part was settled at post #25. I only came in at post #32, where the discussion had moved to the second question, so I assumed your post #41 was in relation to the second question.

OK Alright.