Moment of inertia and angular velocity of a uniform disk

[Sara583]

Homework Statement

Two particles, each of mass m, are attached one to each end of a diameter PQ of a uniform circular disk, of mass 4m, radius a with its centre at O. The system is free to rotate about a horizontal axis through A, a point on PQ such that OA = b as indicated in the diagram below. The system is released from rest when PQ is horizontal.[/B]
Determine the Moment of Inertia of the system about the axis A, in terms of integer constants, a , b and m.
Determine the angular velocity of the disk plus masses when PQ is vertical. Enter your answer in terms of integer constants, a, b and the acceleration due to gravity g.

Homework Equations

I=ma^2
parallel axis equation I=Icentre of mass + md^2

The Attempt at a Solution

I already calculated the moment of inertia using the parallel axis theorem, which gave me 4ma^2+6mb^2, but now I'm not entirely sure how to relate the acceleration to the moment of inertia to find the angular velocity. Any suggestions very welcome.

Thanks

 

[Simon Bridge]

Welcome to PF;
Hint: describe what sort of motion you expect.
See also: Newtons laws and/or conservation of energy.

 

[haruspex]

Two particles, each of mass m, are attached one to each end of a diameter PQ of a uniform circular disk, of mass 4m, radius a with its centre at O. The system is free to rotate about a horizontal axis through A, a point on PQ such that OA = b as indicated in the diagram below. The system is released from rest when PQ is horizontal.
Determine the Moment of Inertia of the system about the axis A, in terms of integer constants, a , b and m.
Determine the angular velocity of the disk plus masses when PQ is vertical. Enter your answer in terms of integer constants, a, b and the acceleration due to gravity g.

I don't see a diagram, and the description does not make clear whether the disk starts in a vertical plane or in a horizontal plane.

I already calculated the moment of inertia using the parallel axis theorem, which gave me 4ma^2+6mb^2

Whichever interpretation I use, I don't get that formula. Please post your working.

 

[Sara583]

here is the diagram for the problem:
My working for part a was)
parallel axis theorem: Icentre of mass + md^2
d= b , total mass = 6m
(ma^2 + ma^2+ 1/2(4m)a^2)+ 6m(b)^2
2ma^2+2ma^2+6mb^2
4ma^2+6mb^2
Hopefully it makes more sense with the diagram now. The program has marked my answer has correct so I assumed I hadn't misapplied the above theorem but if I have, please tell me.

 

[haruspex]

here is the diagram for the problem:
My working for part a was)
parallel axis theorem: Icentre of mass + md^2
d= b , total mass = 6m
(ma^2 + ma^2+ 1/2(4m)a^2)+ 6m(b)^2
2ma^2+2ma^2+6mb^2
4ma^2+6mb^2
Hopefully it makes more sense with the diagram now. The program has marked my answer has correct so I assumed I hadn't misapplied the above theorem but if I have, please tell me.

Ah yes, my mistake.
So have you done the whole question now, or are you still stuck on the second part? Did you try conservation of energy as Simon suggested?

 

[Sara583]

I know that K = 1/2 Iω^2 but I'm still a bit unsure how to calculate the total energy beforehand. Am I missing something?

 

[haruspex]

I know that K = 1/2 Iω^2 but I'm still a bit unsure how to calculate the total energy beforehand. Am I missing something?

The gain in KE has come from a loss of energy somewhere else. What is the nature of that loss and how can you calculate it?

 

[Sara583]

Thank you very much! I did manage to solve it in the end using your hints about energy loss.