# Circular motion in the vertical plane

## Homework Statement

A light rod $AB$ of length $2a$ has a particle $P$ of mass $m$ attached to $B$. The rod is rotating in a vertical plane about a fixed smooth horizontal axis through $A$. Given that the greatest tension in the rod is $\frac {9mg}{8}$, find, to the nearest degree, the angle between the rod and the downward vertical when the speed of $P$ is $\sqrt {\frac {ag}{20}}$.

## Homework Equations

There are two methods that can be applied to this question, as far as my knowledge extends. Newton's Second Law and the conservation of energy.

One, that the tension and weight equal the resultant force, which is the centripetal force.

Second is that the energy at one point equals the energy at another point (of the motion of the object). This is mostly setting equal the kinetic energy at one point with the sum of the KE and PE at another point.

## The Attempt at a Solution

Using Newton's Second Law, I got the equation $T+mgsin\theta=\frac {mv^2}{2a}$. The solution didn't have any, '$cos\theta$' and simply said, '$T-W=F_c$. I don't know why it's subtracted, why there's no angle and which plane I'm resolving in.

Finally, I can't make sense of the energy changes in this question since I don't know which two points to consider. I got nothing near to the second equation in the solution, which is, '$\frac{1}{2} m(\frac {ag}{4})-\frac{1}{2}m(\frac{ag}{20})=mg2a(1-cos\theta)$. I question here why the difference between the two kinetic energies is being found, and also where this, '$1-cos\theta$' is coming from.

I can see there's a gap in my understanding somewhere, so any help would be appreciated.

Thank you

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## Answers and Replies

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Worked it out on my own. The question makes total sense when considering the downwards vertical as the starting point, and any subsequent motion as the, 'final point'. $h=2a(1-cos\theta)$