A simple pendulum has a mass of 0.620 kg and a string of length 1.25 m. It is displaced through an angle of 40.0° to point A and then released from rest. When the mass reaches poin... A simple pendulum has a mass of 0.620 kg and a string of length 1.25 m. It is displaced through an angle of 40.0° to point A and then released from rest. When the mass reaches point B, calculate: (a) the drop in height of the mass, (b) the speed of the mass, and (c) the tension in the string. Read more

Steps to Solve

1. Calculate the vertical height from the pivot to point A

We use trigonometry. The vertical height from the pivot to point A is given by $l \cos(\theta)$, where $l$ is the length of the string and $\theta$ is the angle of displacement.

$$ h_A = l \cos(\theta) = 1.25 \cos(40.0^\circ) $$

2. Calculate the vertical drop in height

The drop in height $h$ is the difference between the length of the string and $h_A$.

$$ h = l - h_A = l - l \cos(\theta) = l(1 - \cos(\theta)) $$ Substituting $l = 1.25 \text{ m}$ and $\theta = 40.0^\circ$: $$ h = 1.25(1 - \cos(40.0^\circ)) $$

3. Calculate the speed of the mass at point B using conservation of energy

The potential energy at point A is converted into kinetic energy at point B. Therefore, $mgh = \frac{1}{2}mv^2$, where $m$ is the mass and $v$ is the speed at point B. Solving for $v$:

$$ v = \sqrt{2gh} = \sqrt{2g \cdot l(1 - \cos(\theta))} $$ Substituting $g = 9.81 \text{ m/s}^2$, $l = 1.25 \text{ m}$, and $\theta = 40.0^\circ$: $$ v = \sqrt{2 \cdot 9.81 \cdot 1.25 (1 - \cos(40.0^\circ))} $$

4. Calculate the tension in the string at point B

At point B, the tension $T$ in the string must support the weight of the mass and provide the centripetal force required for circular motion.

$$ T = mg + \frac{mv^2}{l} $$ Substituting $m = 0.620 \text{ kg}$, $g = 9.81 \text{ m/s}^2$, $v$ from the previous step, and $l = 1.25 \text{ m}$: $$ T = (0.620)(9.81) + \frac{(0.620)v^2}{1.25} $$