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For a vertical circular motion, its standard to solve for tension at the 4 common points on the circle (North, South, East, West)

My question is: Is it possible to solve for tension at other locations?

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Question Setup

A pendulum bob of mass $1.27 \text{kg}$ is supported by a string such that the radius of it’s path is $0.60 \text{m}$. It is moving with velocity of $0.4 \text{m}\text{s}^{-1}$ at angle $a = 30^{\circ}$ south of horizontal.

Can I resolve vertically and horizontally, I.e. vertically : $T\cos \theta =mg$, Horizontally : $T\sin \theta=[\frac{mv^2}{r}]\sin \theta$ Doesn't seem quite right though..

Can anyone advise/explain on how to approach/solve this questions?

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If you resolve the forces vertically and horizontally, it will pose a lot of problems. Just resolve the forces into radial and tangential components, it will work fine. I could not understand the specifications of the problem from the handwriting, so I cannot give a detailed solution. Its not a difficult problem, so i think You will be able to do it yourself :)

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  • $\begingroup$ How to solve by resolving forces into radial and tangential compomemts?): $\endgroup$ – Boomzxc Nov 28 '15 at 3:53
  • $\begingroup$ suppose at any instant, the string makes angle θ with vertical downward direction (measured anticlockwise). Thus the force acting in radial direction will be tension T, and gravity mgcosθ in outward direction. in tangential direction the only force is mgsinθ. Thus the radial component T - mgcosθ provides the centripetal acceleration, and mgsinθ provides the (only) tangential acceleration. Equate these with the given conditions of the problem and you'll get the solution :) $\endgroup$ – Aditya De Saha Nov 28 '15 at 17:00
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There is a fundamental misunderstanding of mechanics in your question. You are not trying to balance forces, since they are not in balance and the mass is accelerating.

There is a radial force that provides the radial, centripetal acceleration. This is the sum of the tension (which acts radially) and a component of the weight. There is also a tangential force which is only provided by a component of the weight and which causes a change in the speed of the weight. This latter component disappears when the weight is at the top and bottom of its circular motion.

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