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The question I'm trying to answer is this: enter image description here

I'm having difficulty with understanding this part of the soloution:

enter image description here

I understand how a formula for $v$, the velocity of the particle, was derived, but I can't understand how a formula for $T$, the tension in the string, was derived. The first thing I can't understand is what component of the weight (radial or transverse) $mg\cos{\theta}$, what it is represented by in the diagram and how it was found. The second thing I cannot understand is why $T-mg\cos{\theta}=\frac{mv^2}{a}$. Just from looking at thing it seems like $\frac{mv^2}{a}$ represents the unbalanced force directed towards the centre of the circle, and that $mg\cos{\theta}$ is in the opposite direction of $T$, making $T-mg\cos{\theta}=\frac{mv^2}{a}$ true. But even if that's true, I still wouldn't know how $mg\cos{\theta}$ would represent the force in the opposite direction of $T$.

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The weight of the mass $m$ is resolved into two components.
The angle between the direction of the weight vector and the radial vector is $\theta$.
It might help if you draw a vector diagram with the weight as the hypotenuse of a right angled triangle as shown below?

enter image description here

A tangential component $mg \sin \theta$ and a radial component $mg \cos \theta$.

The motion is in a circle so the net inward radial force on the mass $T-mg \cos \theta$ produces a centripetal acceleration $\frac {v^2}{a}$.

Then Newton's second law $F=ma$ is used.

$$T-mg \cos \theta = m \frac {v^2}{a}$$

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  • $\begingroup$ I can understand everything you said except from the first bit. I just cant't vosualise the triangle you're talking about. $\endgroup$ – Reinhild Van Rosenú Jul 14 '16 at 13:02
  • $\begingroup$ @ReinhildVanRosenú I have added a diagram. $\endgroup$ – Farcher Jul 14 '16 at 13:14
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At the point P, resolve mg into mgcosx in the radially outward direction and mgsinx in the negative angular direction(if anticlockwise sense is positive). Now, from the reference point of moving P, radially outward forces on it is mgcosx + mv^2/r (second term is the pseudoforce). Radially inward force from the diagram is T. Since there is no radial motion in this frame(or the string wouldve been stretched out), T = mgcosx + mv^2/r The rest follows trivially.

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