The following is an extract from my Further Mechanics 2 book (A level):
So I had a go at the above problem; here is what I attempted:
But the actual answer is apparently this:
Can someone please follow my workings through and point at where I might have gone wrong? (or possibly the mark scheme as this is a first edition copy)
Thanks!
On
With $\theta = \arctan\left(\frac 12\right) = \frac{\pi}{6}$
Calling
$$ \begin{array}{rcll} \delta &=& 2 r \sin\theta\cos\theta&\mbox{Mass distance from rotation axis}\\ \vec H &=&\omega^2\delta m(-1,0)&\mbox{Centripetal force}\\ \vec W &=& m g(0,-1)&\mbox{Weigth force}\\ \vec T &=& t(\sin\theta,\cos\theta)&\mbox{Tension in the rod}\\ \vec R &=& r_0(-\cos\theta,\sin\theta)&\mbox{Sphere reaction force} \end{array} $$
in equilibrium we have
$$ \vec H+\vec W+\vec T+\vec R = \vec 0 $$
or
$$ \left\{ \begin{array}{rcl} -2 m r_0\cos\theta \sin\theta \omega ^2-r \cos\theta+t \sin\theta&=&0 \\ -g m+t \cos\theta+r \sin\theta&=&0 \\ \end{array} \right. $$
solving for $t,r_0$ we have
$$ t=\frac{1}{4} \sqrt{3} m \left(2 g+\omega ^2 r\right)\\ r_0 = \frac{1}{4} m \left(2 g-3 \omega ^2 r\right) $$
The supposed answer with $\sqrt{m}$ cannot be correct. It is not dimensionallly feasible.
Putting the $m$ inside the square root (as in the "actual answer") is obviously wrong.
Your mistake is here: $$ T \sin\theta - R \cos\theta = \frac12 T - \frac{\sqrt3}2 R \neq \frac12 T + \frac{\sqrt3}2 R. $$
You wrote $T \sin\theta - R \cos\theta$ correctly, but later you wrote $\frac12 T + \frac{\sqrt3}2 R,$ which was supposed to be the same thing but was not.
If you had written $\frac12 T - \frac{\sqrt3}2 R$ instead, which is the correct set of substitutions, I think the rest of your calculations would have come out correct.