A particle $P$ of mass m is connected to a fixed point $O$ by a light inextensible string $OP$ of length $r$ and is moving in a vertical circle, centre $O$. At its lowest point, $P$ has speed $U$. When the string makes an angle of $\alpha$ with the downward vertical it encounters a small fixed peg $Q$ where $OQ = \frac{r}{2}$. The string proceeds to wrap itself around the peg, so that $P$ begins to move in a vertical circle with center $Q$ (see diagram). Given that the particle describes a complete circle about $Q$, show that $U^2 \ge gr(\frac{7}{2}-cos\alpha)$
$Note$: The line $OQ$ is solid from $O$ to $Q$ and dotted from $Q$ onward.
Here is what I did:- First I found the difference in height between the lowest point and the highest point as $h=\frac{3r}{2} - \frac{r}{2}cos\alpha$. Then using conservation of energy I said $\frac{1}{2}mU^2= \frac{1}{2}mV^2 +mgh$ Then for the particle at the highest point I used $mg +T = \frac{mv^2}{r}$, made $T$ the subject, made a substitution for $V^2$ using the previous expression and finally set $T\ge0$. I have redone my working a couple of times but was unable to arrive at the given result. Any help would be appreciated
When $P$ is at its highest point above Q, it has a speed enough to make a loop. String tension is at least $0$.
$$mg + T = \frac {mV^2}{r/2} \implies V \ge \sqrt {\frac{gr}2}$$
By conservation of energy, at the lowest point which is $h = \frac32r-\frac12r\cos\alpha$ units lower,
$$\begin{align*} \frac12mU^2 &= \frac12mV^2 + mgh\\ U^2 &\ge \frac{gr}2 +g(3r-r\cos\alpha) \\ &= gr\left(\frac72-\cos\alpha\right) \end{align*}$$