WORKED EXAMPLE

4) A pendulum consists of a brass sphere of mass $5.0 kg$ hanging from a long string (see Figure 5.13).

The sphere is pulled to the side so that it is $0.15 m$ above its lowest position. It is then released.
How fast will it be moving when it passes through the lowest point along its path?
Step 1: Calculate the loss in g.p.e. as the sphere falls from its highest position:
${E_p} = mgh = 5.0 \times 9.81 \times 0.15 = 7.36J$
Step 2: The gain in the sphere’s k.e. is $7.36 J$. We can use this to calculate the sphere’s speed. First, calculate v2, then v:
$\begin{array}{l}
\frac{1}{2}m{v^2} = 7.36\\
\frac{1}{2} \times 5.0 \times {v^2} = 7.36\\
{v^2} = 2 \times \frac{{7.36}}{{5.0}}\\
{v^2} = 2.944\\
v = \sqrt {2.944} \, \approx \,1.72\,m\,{s^{ - 1}}\, \approx \,1.7\,m\,{s^{ - 1}}
\end{array}$
Note that we would obtain the same result in Worked example 4 no matter what the mass of the sphere. This is because both k.e. and g.p.e. depend on mass m. If we write:
$\begin{array}{l}
change\,in\,g.p.e\, = \,change\,in\,k.e\\
mgh\, = \,\frac{1}{2}m{v^2}
\end{array}$
we can cancel m from both sides. Hence:
$\begin{array}{l}
gh\, = \,\frac{{{v^2}}}{2}\\
{v^2} = 2gh\\
v = \sqrt {2gh} 
\end{array}$
The final speed v only depends on g and h. The mass m of the object is irrelevant. This is not surprising; we could use the same equation to calculate the speed of an object falling from height h. An object of small mass gains the same speed as an object of large mass, provided air resistance has no effect.