A Problem About Circular Motion 
A $2$ $\text {kg}$ ball on a string is rotated about a circle of
  radius $10 \text {m}$. The maximum tension allowed in the string is
  $50$ $\text {N}$. 
  
  
*
  
*What is the maximum speed of the ball?
  

May I get your helps for this question? I'm trying to find the correct equation. 
 A: Following Corollary $1$, Prop. $4$, Book I, of Newton's Principia, in this circular motion$$f=\frac{mv^2}{r}$$so we should have$$50=\frac{2v^2}{10}$$from which we get$$v=\sqrt250=15.81\frac{m}{sec}$$
A: This is about rotational motion. Usually there are two cases :
$\textbf{Horizontal motion case}$ : Just use Newton's eq
$$
F_{\text{sent}}=ma_{\text{sent}}=m v^2/r
$$
in this setting the total force is just tension $T$. So,
$$
T_{\text{max}} = m v_{\text{max}}^2/r \implies v_{\text{max}} = \sqrt{rT_{\text{max}}/m}
$$
$\textbf{Vertical motion case}$ : If the motion is vertical, it will be quite different, because now the gravitational force enter the equation. So the tension is varies depend on the position of the ball due to the gravity. 

When this is the case, a little refrection on the free body diagram should convince you that the maximum tension happen when the ball is happen to be at the bottom. So
$$
T_{\text{max}} - mg  = m v_{\text{max}}^2/r \implies v_{\text{max}} = \sqrt{\frac{r}{m}(T_{\text{max}}-mg) }
$$
A: I assume they mean the maximum possible speed of the ball. Use the equation for the centripetal force, $F_c=\frac{mv^2}{r}$, where $v$ is the velocity, $m$ is the mass of the ball and $r$ is the radius. The string exerts this force on the ball, so by newton's third law the ball will exert the same force on the string in outward radial direction. This force can be $50N$ at max, so $v^2$ can be $\frac{50r}{m}$ at max, so $v$ can be at most $\sqrt{\frac{50r}{m}}$. I leave it to you to fill in the numbers. 
