Is Angular Momentum Conserved in Circular Motion?

AI Thread Summary
In circular motion, the angular momentum of a bob suspended from a vertical support changes direction but not magnitude due to the presence of torque from gravity. The tension in the string does not exert torque about the point of suspension, while gravity does, leading to a change in angular momentum. The work done by gravity is zero since it acts perpendicular to the velocity, keeping the kinetic energy constant. Therefore, while the magnitude of angular momentum remains constant, its direction changes, indicating that angular momentum is not conserved in this scenario. The discussion emphasizes the importance of understanding torque and its effects on angular momentum in circular motion.
vijayramakrishnan
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Homework Statement



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[/B]

Homework Equations


torque = rXf
angular momentum = rXp

The Attempt at a Solution


[/B]
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?
 
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Is it rotating in the air? Or is it kept on a surface, like, say, a table?
If it's the former, how can you neglect gravity? Please provide some more information.
 
Qwertywerty said:
Is it rotating in the air? Or is it kept on a surface, like, say, a table?
If it's the former, how can you neglect gravity? Please provide some more information.
sir nothing is given about it,and it does not matter
vijayramakrishnan said:

Homework Statement



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[/B]

Homework Equations


torque = rXf
angular momentum = rXp

The Attempt at a Solution


[/B]
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?
 
vijayramakrishnan said:
sir nothing is given about it,and it does not matter

Qwertywerty said:
Is it rotating in the air? Or is it kept on a surface, like, say, a table?
If it's the former, how can you neglect gravity? Please provide some more information.
sir, i think gravity acts on the system because it is given vertical support.i edited it sir.
 
Ok, so does gravity exert a torque?
 
Qwertywerty said:
Ok, so does gravity exert a torque?
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
 
vijayramakrishnan said:
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
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.
 
haruspex said:
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.
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?
 
vijayramakrishnan said:
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?
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
haruspex said:
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?
i think it is 90 degree.
 
  • #11
vijayramakrishnan said:
i think it is 90 degree.
Right. Do you see how that leads to one of the options?
 
  • #12
haruspex said:
Right. Do you see how that leads to one of the options?
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
vijayramakrishnan said:
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?
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
haruspex said:
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.
haruspex said:
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.
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.
 
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