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a doubt regarding torque

 
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Mar9-12, 09:19 AM   #1
 

a doubt regarding torque


If a body is rotating in a circular orbit then what is the moment of net force acting on it about the axis of rotation?
 
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Mar9-12, 09:22 AM   #2
 
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What do you think?
 
Mar9-12, 09:26 AM   #3
 
0 but how?
 
Mar9-12, 09:26 AM   #4
 
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a doubt regarding torque


Quote by dreamz25 View Post
0 but how?
Which way does the net force act? What's the definition of torque?
 
Mar9-12, 09:35 AM   #5
 
in gravitation i have read that...
the force on a planet towards radius is given by
GMm/r^2
and the force which acts radially outwards is
Mv^2/r
so dunno where the net force acts... !!!:{
and torque = F X R
if the direction of net force is towards center then it makes and angle of
180 degrees which gives ex. torque = 0 as sin 180 = 0 ....
but just tell me about the direction of net external force...!! (Thanks in advance)
 
Mar9-12, 10:02 AM   #6
 
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Quote by dreamz25 View Post
in gravitation i have read that...
the force on a planet towards radius is given by
GMm/r^2
OK.
and the force which acts radially outwards is
Mv^2/r
There is no outward force.
so dunno where the net force acts... !!!:{
The only force acting is gravity.
and torque = F X R
if the direction of net force is towards center then it makes and angle of
180 degrees which gives ex. torque = 0 as sin 180 = 0 ....
Good.
 
Mar10-12, 01:17 AM   #7
 
How then do we derive the velocity of a plannet in circular orbit?????
when we equate both of them...
v = root[GM/R]...!!
 
Mar10-12, 02:07 PM   #8
 
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Quote by dreamz25 View Post
How then do we derive the velocity of a plannet in circular orbit?????
By applying Newton's 2nd law. The only force is gravity. Set that equal to mass X the centripetal acceleration.

No need for any mysterious outward force.
 
Mar10-12, 11:20 PM   #9
 
ok.. so u mean both of them acts towards the centre... Right?
since the force on the particle by the center equals GMm/R^2
and also by Newton's second law of motion, F = ma so, F = m x centripetal acceleration (which is towards the center) = m x v^2/r
and thus we get, v = root[GM/r] ...?
 
Mar11-12, 12:18 AM   #10
 
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Quote by dreamz25 View Post
ok.. so u mean both of them acts towards the centre... Right?
since the force on the particle by the center equals GMm/R^2
and also by Newton's second law of motion, F = ma so, F = m x centripetal acceleration (which is towards the center) = m x v^2/r
and thus we get, v = root[GM/r] ...?
Your math is right, I think you just have a conceptual problem. When you say "both of them acts towards the centre", my question to you is, both of what?

We're not talking about two distinct forces here. Gravity IS the centripetal force in this situation. Centripetal force is always just a requirement for circular motion. It has to be provided by something real, like gravity, or tension in a string. Without something like this to provide (or act as) a centripetal force, there simply won't be any circular motion.
 
Mar11-12, 04:02 AM   #11
 
got ur point.... but i too meant the same...
i m nt differentiating the two force i just meant the different expressions for a single force..
the force between them is GMm/r^2 which also equals mv^2/r (the centripetal force which acts towards the center to keep the body rotating in a circular path) and thus gets the formula derived.....
 
Mar11-12, 04:04 AM   #12
 
Quote by cepheid View Post
Centripetal force is always just a requirement for circular motion. It has to be provided by something real, like gravity, or tension in a string. Without something like this to provide (or act as) a centripetal force, there simply won't be any circular motion.
wonderful lines... thanks..!!
 
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