# Moment of inertia problem

1. Oct 27, 2013

### coldblood

Hi friends the problem is as:
https://fbcdn-sphotos-d-a.akamaihd.net/hphotos-ak-ash4/1379445_1440214886205617_1276077735_n.jpg

While finding moment of inertia of the system about the point from where the rod is fixed, the book says that M.I. of the system will be,

I = Irod + Idisc

I = [1/3m(2R)2] + [m(2R)2]-----------(1)

Sorry in the figure the length of the rod is 2R.

But I think that it should be

I = Irod + Idisc

I = [1/3m(2R)2] + [1/2m(R)2 + m(2R)2] - - - -- - (2)
(Using parallel axis theorem)

Please friends tell why it is so?

While the (2)nd M.I. which I got, the book says that it will be for,
https://fbcdn-sphotos-b-a.akamaihd.net/hphotos-ak-ash4/q71/1378467_1440214882872284_972543142_n.jpg

Please friend help me in this confusion. I will appreciate the help.

2. Oct 28, 2013

### ehild

The rod with a freely rotating disk at the end is not a rigid body, and (2) is true if the disk is fixed to the rod, and rotates together with it, with the same angular velocity.

If some torque acts on the rod, the rod will rotate, but nothing will rotate the disk, as the rod acts with the normal force, and it has zero torque on the disk. The centre of the disk will move but the disk will not rotate with respect to the ground frame of reference, see attached figure. The rotations of the rod and disk are not coupled. If you use the torque equation, the torque is equal to the moment of inertia multiplied by the angular acceleration β. If the body consist of parts, τ=I1β1+I2β2, but the angular acceleration of the disk is zero, so its own angular momentum around its centre does not count.

ehild

#### Attached Files:

• ###### hingeddisk.JPG
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3. Oct 28, 2013

### coldblood

Dear ehild,

When the disc is not free to rotate about its own axis, Can't I assume it as a point mass fixed at the lower end of the rod?
I think I can, cause the CG of the disc will be at that point. So, the moment of inertial of the composite system would be - Irod + Iparticle or disc about the top end of the rod.

So, I = (rod)1/3[m(2R)2] + (disc)m(2R)2

Isn't it correct?

4. Oct 28, 2013

### ehild

You can assume the disk as a point mass at the end of the rod, if it is free to rotate about its own axis. If the rod rotates, the disk may stay unturned with respect to the ground, as no torque acts on it, but rotated with respect to the rod.

When the disk is fixed, all its points have the same angular speed and acceleration as the rod. In this case, its own moment of inertia has to be taken into account.

ehild

5. Oct 28, 2013

### coldblood

Thank you very much ehild.
I got the answer, and got where I was doing mistake.

6. Oct 28, 2013

### ehild

You are welcome

ehild

7. Oct 28, 2013

### Tanya Sharma

Hi ehild…

Could you help me understand and visualize how do we say that the disk is not rotating around the fixed end of the rod when it is free to rotate about its own axis. Any point ,say A or B cover some angle with respect to the fixed end of the rod and the vertical line ,in a given time period .Even though the angles covered would be different for different points i.e different points on the disk move with different angular speeds .Any point on the disk has some angular speed with respect to the fixed end of the rod.

What is the criteria that a rigid body(disk in this case) rotates around a fixed point(fixed end of the rod) ?

Is it that the angular speeds should be same ?

I understand this problem..but somehow I am not convinced that the disk doesn’t rotate around the fixed end of the rod ,when it is not fixed with the rod?

#### Attached Files:

• ###### hingeddisk1.JPG
File size:
10 KB
Views:
68
8. Oct 28, 2013

### ehild

it can rotate, but its rotation is not coupled to that of the rod. If it was in rest with respect to the rod and the rod got a push, the rod would turn around the hinge, but the disk did not get any torque it would not rotate with respect to the ground. But it turned with respect to the rod, as shown in the figure.

ehild