A ball on a string, swung in a level circle, is a neat test of forces at angles and centripetal force. The tension does two jobs at once. Each problem here is one short scene and a chain of questions. Answer them in order: each later question uses what you found before, so a slip early on carries through. Keep full precision between steps and round only at the end.
Key idea
With the string at $\theta$ from the vertical and $r = L\sin\theta$: $$T\cos\theta = mg, \qquad T\sin\theta = \frac{mv^2}{r}.$$ Dividing: $\tan\theta = \dfrac{v^2}{gr}$. The period is $2\pi\sqrt{L\cos\theta/g}$, with no mass in it.
Worked example: A ball on a cord
A $0.3\ \text{kg}$ ball on a $1.0\ \text{m}$ cord circles with the cord at $35^\circ$ from the vertical. Find (1) the tension and (2) the speed.
- (1) $T = \dfrac{(0.3)(9.8)}{\cos 35^\circ} = 3.59\ \text{N}$.
- (2) $r = 1.0\sin 35^\circ = 0.574\ \text{m}$; $v = \sqrt{(9.8)(0.574)\tan 35^\circ} = 1.98\ \text{m/s}$.
Answer: $3.59\ \text{N}$, $1.98\ \text{m/s}$
Common mistake
Using the string length as the radius. The ball circles the vertical line under the pivot, so $r = L\sin\theta$. Check: at a tiny angle the radius should be nearly zero.