Question

A heavy ball with a weight of 100 NN is hung from the ceiling of a lecture hall on a 4.4-mm-long rope. The ball is pulled to one side and released to swing as a pendulum, reaching a speed of 5.7 m/sm/s as it passes through the lowest point

Answer 1

Answer:

175.3 N

Explanation:

The motion of the ball is a uniform circular motion, therefore the net force on it must be equal to the centripetal force.

There are two forces acting on the ball at the lowest point of motion:

- The tension in the string, T , upward

- The weight of the ball, $mg$, downward

The net force (centripetal force) has the same direction as the tension (upward, towards the centre of the circular path), so we can write:

$T-mg=m\frac{v^2}{r}$

where the term on the right is the expression for the centripetal force, and where:

T is the tension in the string

$mg=100 N$ is the weight of the ball

$m=\frac{mg}{g}=\frac{100}{9.8}=10.2 kg$ is the mass of the ball

v = 5.7 m/s is the speed of the ball at the lowest point

r = 4.4 m is the length of the rope, so the radius of the circle

Solving for T, we find the tension in the string:

$T=mg+m\frac{v^2}{r}=(100)+(10.2)\frac{5.7^2}{4.4}=175.3 N$