Answer to Question #116390 in Quantum Mechanics for Foibe Kambala

Question

2. A ball on the end of a string is revolved at uniform rate in vertical circle of radius 72.0
cm. If its speed is 4.00 m/s and mass of 0.3 kg. Calculate the tension
in the string when ball is:
a) At top of its path
b) At bottom of its path

Answer 1

We write the equation of motion of the ball in a circle

$\vec{N}+m\vec{g}=m \vec{a}$

Where

$a=\frac{v^2}{r}$

a) We write the equation of motion for the upper position

$N+mg=\frac{mv^2}{r}$

Then the tension in the string is

$N=\frac{mv^2}{r}-mg=\frac{0.3 \cdot 4^2}{0.72}-0.3 \cdot 9.81=3.724\space{N}$

b) We write the equation of motion for the lower position

$-N+mg=-\frac{mv^2}{r}$

Then the tension in the string is

$N=\frac{mv^2}{r}+mg=\frac{0.3 \cdot 4^2}{0.72}+0.3 \cdot 9.81=9.61\space{N}$