Figure 8-34 shows a thin rod, of length $\mathit{L}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.00}\mathbf{\text{m}}$ and negligible mass, that can pivot about one end to rotate in a vertical circle. A ball of mass $\mathit{m}\mathbf{}\mathbf{=}\mathbf{}\mathbf{5}\mathbf{.00}\mathbf{\text{kg}}$ is attached to the other end. The rod is pulled aside to angle${\mathit{\theta }}_{\mathbf{0}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{30}\mathbf{.0}\mathbf{°}$ and released with initial velocity ${\overrightarrow{\mathbf{v}}}_{\mathbf{0}}\mathbf{=}\mathbf{0}$. As the ball descends to its lowest point, (a) how much work does the gravitational force do on it and (b) what is the change in the gravitational potential energy of the ball–Earth system? (c) If the gravitational potential energy is taken to be zero at the lowest point, what is its value just as the ball is released? (d) Do the magnitudes of the answers to (a) through (c) increase, decrease, or remain the same if angle${\mathit{\theta }}_{\mathbf{0}}$ is increased?

The length, $L=2.00\text{m}$

The mass, $m=5.00\text{kg}$

The rod is pulled aside to angle, ${\theta }_0=30.0°$

Initial velocity, ${\overrightarrow{v}}_0=0$

Free body diagram of the given figure 8-34:

In angle$ACD$,

$\begin{array}{l}\cos \theta =\frac{AC}{L}\\ AC=L\cos \theta \end{array}$

Therefore, from the figure you can write,
$\begin{array}{c}BC=AB-AC\\ =L-L\cos \theta \\ =L(1-\cos \theta )\end{array}$

The work-energy theorem states that the net work done by forces on an object is equal to the change in its kinetic energy.

Figure out the height of the ball using trigonometry and use the relation for calculating the work done.

$\begin{array}{c}\mathbf{\Delta }\mathbf{K}\mathbf{=}\mathbf{W}\\ \mathbf{=}\overrightarrow{\mathbf{F}}\mathbf{\cdot }\overrightarrow{\mathbf{d}}\end{array}$

The main challenge for students in this type of problem seems to be working out the

trigonometry in order to obtain the height of the ball (relative to the low point of the swing) is,

$h=L-L\cos \theta$

For angle$\theta$measured from vertical as shown in free body diagram.

Once this relation (which will not derive here since you have found this to be most easily illustrated on the blackboard) is established, then the principal results of this problem follow from Eq. 7-12 (for$W_g$) and Eq. 8-9 (for$U$).

The vertical component of the displacement vector is downward with magnitude h, so you obtain

$\begin{array}{c}{W}_g=\overrightarrow{F_g}\cdot \overrightarrow{d}\\ =mgh\\ =mgL(1-\cos \theta )\end{array}$

Substitute known values in the above equation.

$\begin{array}{c}{W}_g=\left(5.00\text{kg}\right)(9.80\text{m}/{\text{s}}^{\text{2}})\left(2.00\text{m}\right)(1-\cos 30°)\\ =98\text{J}(1-0.866)\\ =98\text{J}\left(0.134\right)\\ =13.1\text{J}\end{array}$

Calculate the change in the gravitational potential energy of the ball–Earth system as below.

$\begin{array}{c}\Delta U=-W_g\\ =-mgL(1-\cos \theta )\\ =-13.1\text{J}\end{array}$

Calculate the gravitational potential energy just as the ball is released if it is taken to be zero at the lowest point.

Here the gravitational potential energy of earth-ball system is zero at the lowest point. Then gravitational potential energy at the release point is,

$\begin{array}{c}{U}_D=mgL(1-\cos 30°)\\ =13.1\text{J}\end{array}$

As the angle increases, intuitively see that the height $h$ increases (and, less

obviously, from the mathematics, you see that $\cos \theta$decreases so that $(1-\cos \theta )$increases), so the answers to parts (a) and (c) increase, and the absolute value of the answer to part (b) also increases.