Question

A 1.2kg stone is tied to a string and swung in a vertical circle with a radius of 0.75m. The string can withstand a tension of 40.0N. At what maximum speed can a stone move at the bottom of it's path without the string breaking

Answer 1

Hello!

We can begin by summing the forces acting on the stone when it is at the bottom of its trajectory.

Refer to the free-body diagram in the image below for clarification.

We have the force of tension (produced by the string) and the force of gravity acting in opposite directions, so:
$\Sigma F = T - F_g$

The net force is equivalent to the centripetal force experienced by the stone. Recall the equation for centripetal force for uniform circular motion:
$F_c = \frac{mv^2}{r}$

m = mass of object (1.2 kg)
v = velocity of object (? m/s)
r = radius of circle (0.75 m)

The centripetal force is the resultant of the forces of tension and gravity, and points upward (same direction as the tension force) since the tension force is greater.

Therefore:
$\frac{mv^2}{r} = T - Mg$

We can solve the equation for 'v':

$mv^2 = r(T - Mg) \\\\v^2 = \frac{r(T - Mg)}{m}\\\\v = \sqrt{\frac{r(T - Mg)}{m}}$

Plug in values and solve.

$v = \sqrt{\frac{(0.75)(40 - 1.2(9.8))}{1.2}} = \boxed{4.201 \frac{m}{s}}$