The string in Fig. 8-38 is $\mathbf{L}\mathbf{}\mathbf{=}\mathbf{}\mathbf{120}\mathbf{\text{cm}}$long, has a ball attached to one end, and is fixed at its other end. The distance$\mathit{d}$from the fixed end to a fixed peg at point $\mathit{P}$ is $\mathbf{75}\mathbf{.0}\mathbf{\text{cm}}$. When the initially stationary ball is released with the string horizontal as shown, it will swing along the dashed arc. What is its speed when it reaches (a) its lowest point and (b) its highest point after the string catches on the peg?

Length of the string, $\mathrm{L}=120\text{cm}=1.20\text{m}$

The distance from the fixed end to a fixed peg at point $P$, $\mathrm{d}=75.0\text{cm}=0.75\text{m}$

The law of conservation of energy states that energy cannot be created or destroyed - only converted from one form of energy to another. This means that the system always has the same amount of energy unless it is added from outside.

As the string reaches its lowest point, its original potential energy is,

$\mathrm{U}=\mathrm{mgL}$

(Measured relative to the lowest point) is converted into kinetic energy. Thus,

$\mathrm{mgL}=\frac{1}{2}{\mathrm{mv}}^2$

$\mathrm{v}=\sqrt{2\mathrm{gL}}$(1)

Here,

The acceleration due to gravity, $\mathrm{g}=9.80\text{m}/{\text{s}}^{\text{2}}$

Substitute known values in the above equation.

$\begin{array}{c}\mathrm{v}=\sqrt{2(9.80\text{m}/{\text{s}}^{\text{2}})\left(1.20\text{m}\right)}\\ =4.85\text{m}/\text{s}\end{array}$

Hence, the speed when it reaches its lowest point $4.85\text{m}/\text{s}$.

In this case, the total mechanical energy is shared between kinetic $\frac{1}{2}{\mathrm{mv}}_{\mathrm{b}}^2$and potential ${\mathrm{mgy}}_{\mathrm{b}}$. Note that ${\mathrm{y}}_{\mathrm{b}}=2\mathrm{r}$

Where the distance,

$\begin{array}{c}\mathrm{r}=\mathrm{L}-\mathrm{d}\\ =(1.20-0.75)\text{m}\\ =0.450\text{m}\end{array}$

Energy conservation leads to:

$\begin{array}{c}\mathrm{mgL}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{b}}^2+{\mathrm{mgy}}_{\mathrm{b}}\\ \frac{1}{2}{\mathrm{mv}}_{\mathrm{b}}^2=\mathrm{mgL}-{\mathrm{mgy}}_{\mathrm{b}}\\ {\mathrm{v}}_{\mathrm{b}}^2=\frac{2\mathrm{mg}(\mathrm{L}-{\mathrm{y}}_{\mathrm{b}})}{\mathrm{m}}\end{array}$

$\begin{array}{c}{\mathrm{v}}_{\mathrm{b}}=\sqrt{2\mathrm{g}(\mathrm{L}-2\mathrm{r})}\\ =\sqrt{2(9.80\text{m}/{\text{s}}^{\text{2}})(1.20\text{m}-2\left(0.450\text{m}\right))}\\ =\sqrt{(\text{19.6m}/{\text{s}}^{\text{2}})\left(\text{0.3m}\right)}\\ =\sqrt{5.88{\text{m}}^2/{\text{s}}^{\text{2}}}\end{array}$

${\mathrm{v}}_{\mathrm{b}}=2.42\text{m}/\text{s}$

Here, the speed when it reaches its highest point is $2.42\text{m}/\text{s}$.