# Doubt from rotational mechanics

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1. Dec 19, 2015

### ajaysabarish

in newton's laws of rotational motion,
is the net torque calculated only about centre of mass?or is it valid for torque about any axis?

2. Dec 20, 2015

### DaveC426913

For the mass to rotate about any point that is not the centre of mass, there would have to be another external force involved.

As for axes of rotation, a long thin strip of a mass would have a different torque along different axes. Very easy to flip a 2x4 along its long axis (twirling it). Not so easy to flip it along its short axes (end-over-end).

3. Dec 20, 2015

### ajaysabarish

thank you very much sir,but i couldn't really understand your reply.please be little elaborate,i know that newton's law for rotation can be applied to center of mass but in a problem,newton's law of rotation was applied about an hinge.i couldn't understand this,please explain,why is it so?

4. Dec 20, 2015

### JeremyG

It is possible to use the rotational analogue of Newton's law, just be careful of the Moment of Inertia that you use in your equations. The Moment of Inertia must be taken about the rotation axes.

Also, if I'm not wrong, $\tau = I\alpha$ holds only if the axis is fixed. The only exception is if the torque and moment of inertia I are computed about the centre of mass, then the above relation holds if the object is translating with acceleration.

5. Dec 20, 2015

### ajaysabarish

thank you very much for replying,sir.

sir,if we are considering a disc hinged to a rod perpendicular to the plane of disc but it is intersecting the disc at some other than centre,can we still use newton's law for rotation,if yes then what are the axes about which we can apply this law?and about what axis should torque,moment of inertia and angular acceleration be measured?

6. Dec 20, 2015

### JeremyG

Yes you can still use Newton's Law. It would be most useful to consider the torque about the rotation axis, which would passing through the hinge in your question.
So net torque, $\sum\tau = F\times r = I\alpha$ .

In the above equation, everything should be taken with respect to the rotation axis. This includes the moment of inertia I like I was saying in my previous post. So do remember to apply the parallel axis theorem in this case.

7. Dec 20, 2015

### ajaysabarish

thank you sir,it cleared my doubt.

8. Dec 20, 2015

### Staff: Mentor

Not so fast. It you're doing the moment balance about anything other than the center of mass, you need to include pseudo forces (-ma) in the moment balance. It's always safest to do the moment balance about the center of mass.

9. Dec 20, 2015

### ajaysabarish

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