Pendulum Example

A \(\text{2}\) \(\text{kg}\) metal ball is suspended from a rope as a pendulum. If it is released from point A and swings down to the point B (the bottom of its arc):

1. show that the velocity of the ball is independent of its mass,

2. calculate the velocity of the ball at point B.

Step 1: Analyse the question to determine what information is provided

• The mass of the metal ball is \(m = \text{2}\text{ kg}\)

• The change in height going from point A to point B is \(h = \text{0.5}\text{ m}\)

• The ball is released from point A so the velocity at point, \({v}_{A} = \text{0}\text{ m·s$^{-1}$}\).

All quantities are in SI units.

Step 2: Analyse the question to determine what is being asked

• Prove that the velocity is independent of mass.

• Find the velocity of the metal ball at point B.

Step 3: Apply the Law of Conservation of Mechanical Energy to the situation

Since there is no friction, mechanical energy is conserved. Therefore:

\begin{align*} {E}_{M1} & = {E}_{M2} \\ {E}_{P1} + {E}_{K1} & = {E}_{P2} + {E}_{K2} \\ mg{h}_{1} + \frac{1}{2}m{\left({v}_{1}\right)}^{2} & = mg{h}_{2} + \frac{1}{2}m{\left({v}_{2}\right)}^{2} \\ mg{h}_{1} + 0 & = 0 + \frac{1}{2}m{\left({v}_{2}\right)}^{2} \\ mg{h}_{1} & = \frac{1}{2}m{\left({v}_{2}\right)}^{2} \end{align*}

The mass of the ball \(m\) appears on both sides of the equation so it can be eliminated so that the equation becomes:

\begin{align*} g{h}_{1} & = \frac{1}{2}{\left({v}_{2}\right)}^{2} \\ 2g{h}_{1} & = {\left({v}_{2}\right)}^{2} \end{align*}

This proves that the velocity of the ball is independent of its mass. It does not matter what its mass is, it will always have the same velocity when it falls through this height.

Step 4: Calculate the velocity of the ball at point B

We can use the equation above, or do the calculation from “first principles”:

\begin{align*} {\left({v}_{2}\right)}^{2} & = 2g{h}_{1} \\ {\left({v}_{2}\right)}^{2} & = \left(2\right)\left(\text{9.8}\text{ m·s$^{-2}$}\right)\left(\text{0.5}\text{ m}\right) \\ {\left({v}_{2}\right)}^{2} & = \text{9.8}\text{ m$^{2}$·s$^{-2}$} \\ {v}_{2} & = \sqrt{\text{9.8}\text{ m$^{2}$·s$^{-2}$}} \\ {v}_{2} & = \text{3.13}\text{ m·s$^{-1}$} \end{align*}

Alternatively you can do:

\begin{align*} {E}_{K1} + {E}_{P1} & = {E}_{K2} + {E}_{P2} \\ mg{h}_{1} + \frac{1}{2}m{\left({v}_{1}\right)}^{2} & = mg{h}_{2} + \frac{1}{2}m{\left({v}_{2}\right)}^{2} \\ mg{h}_{1} + 0 & = 0 + \frac{1}{2}m{\left({v}_{2}\right)}^{2} \\ {\left({v}_{2}\right)}^{2} & = \frac{2mg{h}_{1}}{m} \\ {\left({v}_{2}\right)}^{2} & = \frac{2\left(\text{2}\text{ kg}\right)\left(\text{9.8}\text{ m·s$^{-2}$}\right)\left(\text{0.5}\text{ m}\right)}{\text{2}\text{ kg}} \\ {v}_{2} & = \sqrt{\text{9.8}\text{ m$^{2}$·s$^{-2}$}} \\ {v}_{2} & = \text{3.13}\text{ m·s$^{-1}$} \end{align*}