Angular Momentum Question concerning a Merry Go Round

[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.