[Physics] Work Done On Circular Motion

Question

I am trying to find the total work done on a ball, $m=0.8kg$, tied to a rope of r=1.6m length and swung in a vertical circle.

I understand the total work done by both the tension in the string and gravity is 0 for a complete circle, but I am having trouble trying to find the work done by gravity when the ball is at its lowest point and at its highest point, work done during a semicircle.

I know that as the ball swings it constantly changes direction and therefore the $\theta$ between $F_g$ and $\Delta s$ changes.

So there is a varying theta that is associated with both $F_g,\Delta S$

I know that $$W_{g}=\int_0^\pi -F_g \cdot dl$$

If $\Delta S = r\theta$

then

$dl=ds=rd\theta$

and $F_g=mg$

$$W_{g}=\int_0^\pi -mgrcos\theta d\theta $$

$$W_{g}=-mgr[sin(\pi)-sin(0)]=0J$$

I don't understand what I am doing wrong the answer is -25.1J. It makes sense that the answer is 0 because when the ball is at the bottom and at the top, $\Delta s$ and $F_g$ are perpendicular to each other and the total work increases until $\frac{\pi}{2}$ then decreases resulting in a 0J.

So what am I doing wrong?

Here is the solution:
I found out that $\Delta s = 2r$ since it started at $y_o=0,y_f=2r=3.2m$
There for $W_g=F_g\Delta s cos\theta=3.2mgcos(180)=-25.1J$

Okay so since the actual solution was so much easier, what exactly did I do wrong with my initial thought of trying to solve it?

And ironically if we do:

$$W_{g}=-2mgr\int_0^{\frac{\pi}{2}} cos\theta d\theta $$

$$W_{g}=-mgr[sin(\frac{\pi}{2})-sin(0)]=-25.1J$$

Answer 1

If $\Delta S = r cos\theta$

then

$dl=ds=rd\theta$

and $F_g=mg$

$$W_{g}=\int_0^\pi mgrcos\theta d\theta $$

If you're taking the angle from the center of the circle (which you are, since you said that $\Delta S = r cos\theta$, then the initial position of the ball is $-R$, since displacement is a vector quantity (and the final position of the ball is on the opposite side of the center of the circle), and hence your angle must vary from $-\frac {\pi}{2}$ to $\frac {\pi}{2}$ instead of $0$ to $\pi$, which gives you the required answer.

It makes sense that the answer is 0 because when the ball is at the bottom and at the top, $\Delta s$ and $F_g$ are perpendicular to each other and the total work increases until $\frac{\pi}{2}$ then decreases resulting in a 0J.

This logic is flawed, because the vertical direction of displacement (which is along the direction of gravity) remains the same (downwards) when the ball is moving from top to bottom, and hence the work done by the Gravitational force is always positive and on the body always negative, during this course.