Simple harmonic motion

  • #1
Kajan thana
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If there is a length thread with a metal ball attached at the end of the thread, and there is a oscilliation.
The restoring force is F=mgsinθ, my question is can we consider this as a centripetal force and link it to this equation: mv^2/r.
 

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  • #2
Orodruin
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If there is a length thread with a metal ball attached at the end of the thread, and there is a oscilliation.
The restoring force is F=mgsinθ, my question is can we consider this as a centripetal force and link it to this equation: mv^2/r.
Yes, but only for small ##\theta## and for circular orbits. For large ##\theta## you are no longer getting a good approximation of the motion by approximating it with motion in a plane and for non-circular orbits ##F = mv^2/r## does not apply.
 
  • #3
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If there is a length thread with a metal ball attached at the end of the thread, and there is a oscilliation.
The restoring force is F=mgsinθ, my question is can we consider this as a centripetal force and link it to this equation: mv^2/r.
Well what kind of motion gives a centripetal force given by the constant ##mv^2/r## ?
 
  • #4
Orodruin
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No, this force is the tangential force. The force that acts as a centripetal force is the tension in the string and this is equal to [itex] mg cos \theta [/itex]. Note that the speed is not constant so that the value of v keeps changing.
It is the centripetal force in the horizontal plane as long as ##\theta## is small enough for this approximation to be valid. At least this is how I interpreted the question.

Edit: Also, in the planar pendulum case, the tension is not just ##mg\cos(\theta)##, but depends on the velocity as well. You will find ##T - mg \cos(\theta) = mv^2/r##.
 
  • #5
nasu
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Can you describe how you see the setup so that mg sin(θ) is centripetal force? With the appropriate approximation, of course.
 
  • #6
nrqed
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It is the centripetal force in the horizontal plane as long as ##\theta## is small enough for this approximation to be valid. At least this is how I interpreted the question.

Edit: Also, in the planar pendulum case, the tension is not just ##mg\cos(\theta)##, but depends on the velocity as well. You will find ##T - mg \cos(\theta) = mv^2/r##.
Yes, you are absolutely right that the centripetal force is not mg cos theta, I said something stupid there. However, the centripetal force is not equal to mg sin theta, the centripetal force is equal to the the net force towards the center of the motion, which is T - mg cos theta.
 
  • #7
Orodruin
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However, the centripetal force is not equal to mg sin theta, the centripetal force is equal to the the net force towards the center of the motion, which is T - mg cos theta.
Did you read my interpretation of the problem? In that interpretation ##mg \sin(\theta)## is the centripetal force! It is just with a different centre than the one you are thinking of.

Can you describe how you see the setup so that mg sin(θ) is centripetal force? With the appropriate approximation, of course.
I was thinking of a spherical pendulum, i.e., letting the pendulum swing in both directions and not only in a plane, as there is nothing in the OP that restricts the motion to a plane. For small angles, the curvature of the sphere that the pendulum is restricted to is well approximated by a flat surface and the restoring force within this plane is ##mg \sin(\theta)## directed towards the centre.

You can also see this by deriving the full equations of motion for the spherical pendulum and linearising them around ##\theta = 0##. You will find a linearised system equivalent to a two-dimensional centripetal problem with a centripetal force ##mg \theta = mgr/\ell##, where ##r## is the distance from the low point and ##\ell## is the length of the pendulum.
 
  • #8
nrqed
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Did you read my interpretation of the problem? In that interpretation ##mg \sin(\theta)## is the centripetal force! It is just with a different centre than the one you are thinking of.


I was thinking of a spherical pendulum, i.e., letting the pendulum swing in both directions and not only in a plane, as there is nothing in the OP that restricts the motion to a plane. For small angles, the curvature of the sphere that the pendulum is restricted to is well approximated by a flat surface and the restoring force within this plane is ##mg \sin(\theta)## directed towards the centre.

You can also see this by deriving the full equations of motion for the spherical pendulum and linearising them around ##\theta = 0##. You will find a linearised system equivalent to a two-dimensional centripetal problem with a centripetal force ##mg \theta = mgr/\ell##, where ##r## is the distance from the low point and ##\ell## is the length of the pendulum.
I was thinking about the usual set up for a pendulum, with the mass oscillating in a vertical plane. What made me think of my set up instead is that the OP talked about a restoring force so that, to me, this implied oscillation around the equilibrium position. I guess we have to ask him/her for more details. If you consider a general motion with theta not constant and with the oscillation not in a purely horizontal plane, it does not seem to me that we can talk about a centripetal force since the motion is not circular.
 
  • #9
Orodruin
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If you consider a general motion with theta not constant and with the oscillation not in a purely horizontal plane, it does not seem to me that we can talk about a centripetal force since the motion is not circular.
As long as you stay at small angles, the curvature of the sphere is negligible. For small angles the problem is exactly equivalent to a two-dimensional harmonic oscillator.
 
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