Figure 8-31 shows a ball with mass $\mathit{m}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.341}\mathbf{\text{kg}}$attached to the end of a thin rod with length$\mathit{L}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.452}\mathbf{\text{m}}$and negligible mass. The other end of the rod is pivoted so that the ball can move in a vertical circle. The rod is held horizontally as shown and then given enough of a downward push to cause the ball to swing down and around and just reach the vertically up position, with zero speed there. How much work is done on the ball by the gravitational force from the initial point to (a) the lowest point (b) the highest point (c) the point on the right level with the initial point?If the gravitational potential energy of the ball-Earth system is taken to be zero at the initial point, what is it when the ball reaches (d) the lowest point (e) the highest point, and (f) the point on the right level with the initial point? (g) Suppose the rod were pushed harder so that the ball passed through the highest point with a nonzero speed. Would$\mathit{\Delta }{\mathit{U}}_{\mathbf{g}}$from the lowest point to the highest point then be greater than, less than, or the same as it was when the ball stopped at the highest point?

i) Mass of ball$m=0.341\text{\hspace{0.17em}kg}$

ii) Length of thin rod$L=0.452\text{m}$

iii) Gravitational acceleration$g=9.8{\text{m/s}}^{\text{2}}$

By using the concept of potential energy, we can find gravitational work. Gravitational work is nothing but the change in potential energy due the gravitational force.

Formula:

Gravitational potential energy is given by formula

$\begin{array}{c}U=W_g\\ =mgh\end{array}$

Workdepends on theinitial and final position. The only force that does work on the ball is the force of gravity; the force of the rod is perpendicular to the path of the ball, and so, does no work.Going from its initial position to the lowest point on its path, the ball moves vertically through a distance equal to the length Lof the rod, so the work done by the force of gravity is

$W_g=mgL$

$\Rightarrow W_g=0.341\times 9.80\times 0.452$

$\Rightarrow W_g=1.51\text{\hspace{0.17em}J}$

In going from its initial position to the highest point on its path, the ball moves vertically through a distance equal to L, but this time the displacement is upward, opposite to the direction of the force of gravity. The work done by the force of gravity is

$W_g=-mgL$

$\Rightarrow W_g=-0.341\times 9.80\times 0.452$

$\Rightarrow W_g=-1.51\text{\hspace{0.17em}J}$

Thefinal position of the ball is at the same height as its initial position. The displacement is horizontal, perpendicular to the force of gravity. The force of gravity does not work during this displacement i.e.

$W_g=0\text{\hspace{0.17em}J}$

The force of gravity is conservative. The change in the gravitational potential energy of the Ball-Earth system is negative of the work done by gravity

As ball goes to final point

$\Delta U=-mgL$

$\Rightarrow \Delta U=-0.341\times 9.80\times 0.452$

$\Rightarrow \Delta U=-1.51\text{\hspace{0.17em}J}$

For highest point

$\Delta U=+mgL$

$\Rightarrow \Delta U=+0.341\times 9.80\times 0.452$

$\Rightarrow \Delta U=+1.51\text{\hspace{0.17em}J}$

Change in potential energy at same height is zero i.e.

$\Delta U=mg(L-L)$

$\Rightarrow \Delta U=0\text{\hspace{0.17em}J}$

The change in the gravitational potential energy depends on the initial and final positions of the ball, not on its speed. The change in the potential energy is the samesince the initial and final positions are the same.