Question

A 5.0 kg Foucalt Pendulum swings at the end of a 4.0 m long cable. The pendulum is released from a height of 1.5 m above the lowest position of its swing. What is the maximum tension in the cable?

Answer 1

Hi there!

We can begin by solving for the pendulum's velocity at the bottom of its trajectory using the work-energy theorem.

Recall:
$E_i = E_f$

Initially, we just have Potential Energy. At the bottom, there is just Kinetic Energy.

$PE = KE$

Working equation:
$\large\boxed{mgh = \frac{1}{2}mv^2}$

Rearrange to solve for velocity:
$gh = \frac{1}{2}v^2\\\\v = \sqrt{2gh}\\\\v = \sqrt{2(9.8)(1.5)} = 5.42 \frac{m}{s}$

Now, we can do a summation of forces:
$\Sigma F = T - W$

The net force is the centripetal force, so:
$\frac{mv^2}{r} = T - W$

Rearrange to solve for tension:
$T = \frac{mv^2}{r} + W\\\\T = \frac{5(5.42^2)}{4} + 5(9.8) = \boxed{85.75 N}$