# Rotational motion.

1. Mar 2, 2006

### vaishakh

This is the doubt that troubled me when the chapter of rotational motion started. If torque is applied on a body for sometime and then the torque is stopped, then will the body continue to rotate for ever?

Yes - coz the rotational Kinetic energy Iw^2 must be conserved since no mommentum is transferred to the system.

No - coz there is no force for the centre of mass of the body to accelerate. I mean in the given condition it could be that rotational axis is not a centroidal one.

2. Mar 2, 2006

### arildno

Remember that a rigid body can be regarded as rotating about ANY of its constituent points with the the same angular velocity as about any other point.
Thus, when the applied torque stops, and the rotation axis does not go through C.M (i.e, non-zero C.M-velocity), it just means that your body will have a non-zero C.M (constant) velocity in addition to its constant angular velocity.
Its kinetic energy will be: $$\frac{1}{2}(mv_{C.M}^{2}+I_{C.M}\omega^{2})$$

3. Mar 2, 2006

### vaishakh

Then doesn't that oppose Newton's First law of motion as the body is accelerating and no force being applied on it?

4. Mar 2, 2006

### arildno

Whenever did it's C.M ever accelerate after the torque stops?

5. Mar 2, 2006

### vaishakh

I am talking about the radial acceleration. But now I pointed out the defect. Yes your points helped me a lot. The axis which could be external and fixed applies a normal force on the object which leads it to be in rotation. If such an axis does not exist then the object would decide to rotate on centre. Now another doubt -

What would happen if such a fixed axis on which the body is rotating is removed, alongwith stopping to apply torque?

6. Mar 2, 2006

### arildno

Remember that the concept of "rotation axis" isn't a fundamental concept in rigid body mechanics; in contrast to angular velocity and C.M velocity (not to mention the concepts of angular&linear momentum).

If no external forces acts upon a body, then its angular momentum with respect to its C.M remains constant, and we should perhaps better write the rotational part of the body's kinetic energy as $\frac{1}{2}{\vec{S}}\cdot\vec{\omega}$ where $\vec{S}$ denotes the angular momentum
Thus, since the C.M. velocity is constant as well, it follows that conservation of energy yields $\vec{S}\cdot\frac{d\vec{\omega}}{dt}=0$,