Question

A student swings a 0.35 kg rock on a 1.25 m string above their head. The rock goes through 1 complete revolution in 0.665 s. What is the tension in the string?

Answer 1

Answer:

39.0 N

Explanation:

Since the rock is in uniform circular motion, the tension in the string provides the centripetal force required to keep the rock in circular motion.

So we can write:

$T=m\omega^2 r$

where

T is the tension in the string

m = 0.35 kg is the mass of the rock

r = 1.25 m is the radius of the circle (the length of the string)

$\omega$ is the angular velocity of the rock

Here the rock covers 1 revolution in a time of

t = 0.665 s

1 revolution corresponds to an angle of $2\pi$ rad, so the angular velocity is:

$\omega=\frac{2\pi}{0.665 s}=9.44 rad/s$

Therefore substituting into the previous equation, we can find the centripetal force, which is the tension in the string:

$T=(0.35)(9.44)^2(1.25)=39.0 N$