TomHart said: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.
Do you have the shape of a flat disk?PhysicsA1 said:I= 520 kg*m (Due to taking the initial mass (200) and adding 60
I do not understand this step. What calculation is this?PhysicsA1 said:40=520a
haruspex said:Are you the shape of a flat disk?
That's the merry-go-round, not you, the 60kg passenger.PhysicsA1 said:yes. its a playground merry go round
haruspex said:That's the merry-go-round, not you, the 60kg passenger.
Yes, I read and understood the question.PhysicsA1 said: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 said: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.
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.PhysicsA1 said:So it will only be effecting the side of which i jump on. How in return will that actually be calculated then?
haruspex said: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?
See http://hyperphysics.phy-astr.gsu.edu/hbase/mi.htmlPhysicsA1 said:I honestly have no clue
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.Caio Graco said: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?
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 said: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?
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 said: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.
OK Alright.haruspex said: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.