# Kinetic energy of rotation and parallel axis theorem problem

1. Dec 2, 2017

### LuigiAM

1. The problem statement, all variables and given/known data

A circular disc of radius 25 cm and mass 0.5 kg is revolving in its plane with an angular velocity of 4 radians per second. Find A) its kinetic energy of rotation, and B) its new angular velocity if a mass of 10 kg is suddenly fixed on the rim of the disc.

2. Relevant equations
Ek = 1/2 I ω2

Moment of inertia for disc = I = 1/2 mr2

Parallel axis theorem: Io = 1/2 mr2 + mr

Conservation of angular momentum: Li = Lf

3. The attempt at a solution

A)
I = 1/2 mr2
I = 1/2 (0.5 kg) (0.25 m)2 = 0.015625 kg m2

Kinetic energy of rotation = = 1/2 I ω2
Kinetic energy of rotation = 1/2 (0.015625)(4)2 = 0.125 J

B)
The new mass shifts the axis of rotation from the center point of the disc to a point O on the rim of the disc. The distance between the two points is equal to the radius (0.25 m). Parallel axis theorem applies:

Io = moment of inertia on the new point on the rim of the disc
Io = 1/2 mr2 + mr
Io = 1/2(0.5 kg)(0.25 m)2 + (0.5 kg)(0.25 m)
Io = 0.140625 kg m2

Since there are no external torque acting of the disc, conservation of angular momentum applies. So, initial angular momentum is equal to final angular momentum.

I * ωi = Io * ωf
(0.0156 kg m2)(4 rad/s) = (0.140625 kg m2) ωf

The new angular velocity is 0.44 rad /s

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I did my best to solve the problem as well as I could. Is my reasoning correct?

One thing I am not sure at all about is the effect of the mass of 10kg that was used to basically fix a new point of rotation on the disc. The way I understand it is that all that new mass does is just shift the axis of rotation to a new point, so we apply the parallel axis theorem to get the new moment of inertia. So the calculations would be the same if the mass was 20kg or 50kg, etc.

I appreciate any help!

Thanks

2. Dec 2, 2017

### Abhishek kumar

How can you say that new mass will shift the axis of rotation if there is no any external force on it and one error in calculation of new moment of inertia you added mr but it will be mr^2

3. Dec 2, 2017

### LuigiAM

Hi,

Thanks for the response

Oh thanks, yes I see the error in calculation.

For the rest: I assume the new mass will shift the axis of rotation because:

The way the problem is written, the disc is revolving in its plane so it is rotating about its center. Angular velocity is constant so it's just an equilibrium rotation.

If we fix a mass to the edge of the rim, isn't it like nailing a wooden bar to a wall? When we nail a wooden bar to a wall, the bar will tend rotate about the nail instead of about its center of mass.

Is it mistaken to assume this is the same kind of problem but with a disc instead of a bar? My first reflex when reading it was that the big weight would just stop the disc altogether, but it says nothing about the size of the weight or about friction, etc.

4. Dec 2, 2017

### Abhishek kumar

Either bar or disc the concept will be same.if axis of rotation will changed then how can you apply conservation of momentum.

5. Dec 2, 2017

### LuigiAM

Hi,

Thanks for the help.

My notes say that if the sum of the external torques on a rotating body is zero, then the total angular momentum remains constant.

My understanding is that the new weight that is fixed to the rim simply changes the axis of rotation but is not in itself a torque since it doesn't act in or against the direction of rotation. It changes the moment of inertia, which leads to a change in angular velocity, but it does not give an angular acceleration.

6. Dec 2, 2017

Correct

7. Dec 2, 2017

Yeah! Thanks

8. Dec 2, 2017

### haruspex

Only if the attached mass is infinite, which it isn't.
It is given as 10kg.