Question

A pendulum is made of a small sphere of mass 0.250 kg attached to a lightweight string 1.20 m in length. As the pendulum swings back and forth, the maximum angle that the string makes with the vertical is 34.0∘. Friction can be ignored. At the low point of the sphere's trajectory, what is the tension in the string?

Answer 1

Answer:

0.833 N

Explanation:

Formula for Kinetic Energy $E_k = \frac{mv^2}{2}$

Formula for Potential Energy $E_p = mgy$

First we need to find the vertical distance between the maximum-angle position and the pendulum lowest point:

Using the swinging point as the reference, the vertical distance from the maximum-angle (34 degree) position to the swinging point is:

$L * cos(34^o) = 1.2cos(34^o) = 1.2*0.83 = 0.995 \approx 1 m$

At the lowest position, pendulum is at string length to the swinging point, which is 1.2 m. Therefore, the vertical distance between the maximum-angle position and the pendulum lowest point would be

y = 1.2 - 1 = 0.2 m.

As the pendulum is traveling from the maximum-angle position to the lowest point position, its potential energy would be converted to the kinetic energy.

By law of energy conservation:

$E_k = E_p$

$\frac{mv^2}{2} = mgy$

$v^2 = 2gy$

$v = \sqrt{2gy}$

Substitute $g = 10 m/s^2$ and y = 0.2 m:

$v = \sqrt{2 * 10 * 0.2} = \sqrt{4} = 2 m/s$

At lowest point, pendulum would generate centripetal tension force on the string:

$F = m\frac{v^2}{L}$

We can substitute mass m = 0.25, rotation radius L = 1.2 m and v = 2 m/s:

$F = 0.25\frac{2^2}{1.2} = 0.833 N$