Kinetic energy of rotation and parallel axis theorem problem

[LuigiAM]

Homework Statement

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.

Homework Equations

E_k = 1/2 I ω²

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

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

Conservation of angular momentum: L_i = L_f

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:

I_o = moment of inertia on the new point on the rim of the disc
I_o = 1/2 mr² + mr
I_o = 1/2(0.5 kg)(0.25 m)² + (0.5 kg)(0.25 m)
I_o = 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 * ω_i = I_o * ω_f
(0.0156 kg m²)(4 rad/s) = (0.140625 kg m²) ω_f
ω_f = 0.44 rad /s

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

 

[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

 

[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.

 

[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.

 

[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.

 

[Abhishek kumar]

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.

Correct

 

[LuigiAM]

Yeah! Thanks

 

[haruspex]

If we fix a mass to the edge of the rim, isn't it like nailing a wooden bar to a wall?

Only if the attached mass is infinite, which it isn't.

it says nothing about the size of the weight

It is given as 10kg.

 

[KMconcordia]

So for part B is the correct formula, Io = 1/2 mr^2 + mr^2

Why is this?

Sorry so late to this post.

 

[haruspex]

So for part B is the correct formula, Io = 1/2 mr^2 + mr^2

No.
There are a couple of ways to approach this problem.
You can find the mass centre of the system then use that angular momentum will be conserved about that point. I tend to avoid finding mass centres of systems because there is usually an easier way, but in this case it might be simplest.
Alternatively, you can take conservation of angular momentum about any fixed point - the initial position of either the centre of the disc or the point where the mass is attached, say. This way, you can just treat each body as contributing to angular momentum, but you will also need to use conservation of linear momentum. E.g. if you take the initial position of the disc's centre as axis then the subsequent linear motion of the attached particle needs to be related to the new angular rotation rate.

 

[KMconcordia]

My problem Is slightly different, would you say this is correct? Sorry I am just very confused.

 

[haruspex]

My problem Is slightly different, would you say this is correct? Sorry I am just very confused.

You should really open your own thread, but it certainly is the same as the original problem in this thread.
Your calculation of the new MoI is in respect of an axis at the perimeter of the disc. That would be fine if that point were the new centre of rotation, but it will not be. The mass of the attached object must feature in the answer. E.g. if the attached mass is very small there will be little change to the rotation of the disk.