Angular rotation where radius is unknown

Question

A light inextensible string AB of length 90 cm has a particle of mass 600 g fastened to it at a point C. The ends A and B are attached to two fixed points in the same vertical line as each other, with A 60 cm above B. The particle moves on a horizontal circle at a constant angular speed of 5 rad/s with both parts of the string taut and CB horizontal. Find the tensions in the two parts of the string.

Resolving vertically $T_1sin\theta=mg=0.6g$

Resolving horizontally $T_2+T_1cos\theta=mr\omega^2=15r$

How can I find the radius given the total string length of 90cm?

unknown 'r'

Answer 1

If you know the distances $AC$ and $CB$ then you can form the triangle $\triangle ABC,$ and the radius you want is the distance of $C$ from the side $AB.$

Since the problem statement says the total length of the string is $90$ and both parts are taut, you have $AC + CB = 90.$ You also have $CB$ horizontal, which tells you that $CB = r$ and that $\triangle ABC$ is a right triangle. (Remember what the Pythagorean Theorem says about a right triangle.)

Can you draw the triangle and find the answer from here?

Answer 2

\begin{align} T_1\sin\theta &=mg\\ T_2+T_1\cos\theta &=mr\omega^2 \end{align} that is \begin{align} T_1&=\frac{mg}{\sin\theta }\\ T_2+mg\cot\theta &=mr\omega^2 \end{align}

Putting $AC=\ell$, $CB=r$, $\ell+r=90\textrm{ cm}$, $AB=h=60\textrm{ cm}$, from $h^2+r^2=\ell^2=(90-r)^2$ we have $r=25\textrm{ cm}$ and $\ell=65\textrm{ cm}$

So we have $\sin\theta=\frac{h}{\ell}$, $\cot\theta=\frac{r}{h}$

\begin{align} T_1&=\frac{mg}{\sin\theta }=\frac{mg\ell}{h}=6,37\mathrm{ N}\\ T_2&=mr\omega^2-\frac{mgr}{h}=1,3\mathrm{ N} \end{align}