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1. Mar 13, 2016

### vijayramakrishnan

1. The problem statement, all variables and given/known data

A bob of mass m attached to an inextensible string of length l is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed ω rad/s about the vertical. About the point of suspension :

(1) angular momentum changes in direction but not in magnitude.

(2) angular momentum changes both in direction and magnitude.

(3) angular momentum is conserved.

(4) angular momentum changes in magnitude but not in direction

2. Relevant equations
torque = rXf
angular momentum = rXp

3. The attempt at a solution

i know that answer is a and how it came but i feel it is answer c,because there is only one force tension, and the torque produced by it is zero about point of suspension as r is parallel to f,so shouldn't momentum be conserved?

Last edited: Mar 13, 2016
2. Mar 13, 2016

### Qwertywerty

Is it rotating in the air? Or is it kept on a surface, like, say, a table?

3. Mar 13, 2016

### vijayramakrishnan

sir nothing is given about it,and it does not matter

4. Mar 13, 2016

### vijayramakrishnan

sir, i think gravity acts on the system because it is given vertical support.i edited it sir.

5. Mar 13, 2016

### Qwertywerty

Ok, so does gravity exert a torque?

6. Mar 14, 2016

### vijayramakrishnan

sir it is not given,but i think it does ,else in a gravity free space how can something rotate in when the string is suspended from a vertical support

7. Mar 14, 2016

### haruspex

It doesn't need to be given. Gravity exerts a force. A force has a magnitude, a direction and a line of action. If you pick some point not in that line of action, the force has a torque about that point.
In the present case, the mass is not directly below the point of suspension, therefore the force of gravity on the mass has a torque about the point of suspension.

8. Mar 14, 2016

### vijayramakrishnan

thank you very much sir for replying,so tension doesn't exert a torque but gravity does,so momentum is not conserved am i right sir?

9. Mar 14, 2016

### haruspex

In respect of the point of suspension, yes. (You should always make the reference point/axis clear when discussing moments etc.)
What is the angle between:
- the angular momentum vector through the point of suspension, at some instant, and
- the torque due to gravity about the point of suspension?

10. Mar 15, 2016

### vijayramakrishnan

i think it is 90 degree.

11. Mar 15, 2016

### haruspex

Right. Do you see how that leads to one of the options?

12. Mar 15, 2016

### vijayramakrishnan

i think here work done by gravity is zero since gravity is perpendicular to velocity,so kinetic energy of the particle is constant so magnitude of angular momentum is constant but direction changes because there is a torque by gravity,am i right sir?

13. Mar 15, 2016

### haruspex

Yes, that's all correct. But what I was thinking of is that torque produces change of angular momentum, $\vec \tau=\dot{\vec L}$. If the torque is perpendicular to the angular momentum then the dot product is zero, so $\vec L.\dot{\vec L}=0$. Integrating, $|\vec L|^2=c$, for some constant c.

14. Mar 15, 2016

### vijayramakrishnan

thank you very much sir for replying but can we integrate like that because i have not been taught integrating dot product of vectors,we write it as abcoθ and integrate each term.