# Homework Help: Is moment of inertia only for rotating objects?

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1. Feb 29, 2016

### Titan97

1. The problem statement, all variables and given/known data
A rod attached to a ceiling at one end and a disc on the other end is performing SHM. In case (1) the disc cannot rotate. In case (2) the disc can rotate about its centre. Compare the restoring torque and angular frequency in both cases.

2. Relevant equations
$$T=2\pi\sqrt{\frac{I}{MgL}}$$

3. The attempt at a solution
I think the restoring torques are same. But my friend says there will be hinge forces that will affect the restoring torque in case (2) because of rotation of disc which causes centripetal force on disc.

In the second question, again we got different answers. Using the formula specified above, the frequency will be different because of an additional moment of inertia in case (2). My friend told that even if the disc is not rotating in case (1), it's moment of inertia is still $\frac{Mr^2}{2}$. So he says the the frequency is same.
I am confused now. Is moment of inertia only for rotating objects?

(Both my answers match with the given solution)

2. Feb 29, 2016

### BvU

Friend has things mixed up: in case (2) disk can rotate and if there is no friction at the axis of rotation, disk does not rotate (since there is nothing that can make it do so). It behaves as a weight concentrated at its center.
So do I. You probably even have an expression for it from which that can be seen.
Difficult to stay consistent here, eh ? In case (1) the disk rotates. In case (2) it does not. See above. Anyway, friend is wrong.

3. Feb 29, 2016

### Titan97

So moment of inertia is only for rotating objects? Also, can you elaborate on hinge forces that act when the disc rotates? The hinge force can have a component along the perpendicular to the rod which can exert torque.

4. Feb 29, 2016

### BvU

Case (1) requires the disk to rotate: pins or glue at the red dots exercise torque.

In case (2) that is not so

And I don't know about hinge forces. What's that ?

5. Feb 29, 2016

### haruspex

The wording of the question is a little unfortunate. I assume case 1 means the disc cannot rotate relative to the rod, and therefore it does rotate relative to the rod's pivot.
I believe a hinge force refers to the radial force when an object rotates about a hinge, i.e. the tension providing the centripetal force. If so, it has no relevance to this question.

6. Mar 1, 2016

### Titan97

@haruspex wont the moment of inertia be same in both cases?

7. Mar 1, 2016

### BvU

No. That's the whole point of the exercise !

8. Mar 1, 2016

### Titan97

@BvU those diagrams are clear. Nice job. In the second case, the disc is not rotating. But even if it does not raotate, the moment of inertia of a disc is $\frac{mr^2}{2}$

9. Mar 1, 2016

### haruspex

Not sure what point you are making. Yes, that's the moment of inertia of a disk about its centre, but if it is not going to rotate about its centre it might as well be a point mass.

10. Mar 1, 2016

### Titan97

So moment of inertia of an object that does not rotate about any point is zero?

11. Mar 1, 2016

### haruspex

No, that's not what I wrote. The moment of inertia (about a given axis) is an intrinsic property, regardless of the state of motion, somewhat like mass. But it is not of interest in an arrangement in which it never actually rotates. In such an arrangement, it might as well be a point mass.

12. Mar 3, 2016

### SammyS

Staff Emeritus
Moment of Inertia is useful for analyzing the motion of a rigid body. All portions of a rigid body have a single uniform value for angular velocity at all times.

In case 1, you do have a single rigid body.

In case 2 you have two rigid bodies, one of which (the disc) does not rotate. The disc does indeed have a Moment of Inertia, but its rotational energy and angular momentum don't change. (As I read it they're both zero.) In the end the disc can be treated as a point mass as haruspex pointed out.