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For a particle attached to a rope moving along a circle that has the length of rope as radius, the tension provides centripetal force and work done by tension is zero since velocity of particle is perpendicular to tension at any instant. But I am thinking about this case: top view

Here, a particle is tied to a vertical cylinder and is given a velocity perpendicular to rope. My doubt is the work done by tension. The length of rope decreases and the particle moves towards P which is along the direction of tension. So the work done is non zero. Am I correct? (Let the particle move in a frictionless surface)

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No it is still zero.

Take $\theta$ to be the angle between the point at the bottom of the cylinder, C and P. Take the radius of the cylinder to be $R$ and max length of rope $L$.

The vector from C to P is $$r_{CP} = R(\sin \theta,-\cos\theta).$$

The vector from P to the particle is $$r_{T}=(L-R\theta)(\cos\theta,\sin\theta).$$

So the position vector of the particle is $r = r_{CP}+r_{T}.$ The tangent vector to the path of the particle is $$\frac{dr}{d\theta}=(L-R\theta)(-\sin\theta,\cos\theta)$$ which is perpendicular to the direction of tension $r_T$,

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  • $\begingroup$ So you are saying the particle has velocity perpendicular to tension. $\endgroup$ – Aditya Dev Aug 30 '15 at 4:54
  • $\begingroup$ How did you get the tangent vector? I am not getting the same answer. $\endgroup$ – Aditya Dev Aug 30 '15 at 4:58
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    $\begingroup$ Use the product rule on $dr_T/d\theta$. The term from taking the derivative of $R\theta$ cancels with the term from $dr_{CP}/d\theta$ $\endgroup$ – octonion Aug 30 '15 at 5:02
  • $\begingroup$ This is my doubt. If you differentiate position vector of particle, shouldn't you get a vector perpendicular to that position vector? $\endgroup$ – Aditya Dev Aug 30 '15 at 5:24
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    $\begingroup$ What you might be thinking of is if the position vector does not change its length, then you can show the tangent vector must be perpendicular. $\endgroup$ – octonion Aug 30 '15 at 14:25

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