homework 5 begin date: 8/17/2025 11:00:00 pm due date: 9/30/2025 11:59:00 pm end date: 9/30/2025 11:59:00 pm…

Explanation:

Step1: Apply conservation of mechanical energy

The initial potential - energy of the ball is $U = mgh$. At the lowest point of the semi - circular tube, this potential energy is converted into kinetic energy $K=\frac{1}{2}mv^{2}$ and some remaining potential energy $U' = mgr$. So, $mgh=mgr+\frac{1}{2}mv^{2}$. We can solve for $v^{2}$: $v^{2}=2g(h - r)$

Step2: Analyze the forces at the lowest point

At the lowest point of the semi - circular tube, the net force acting on the ball is $F_{net}=N - mg$, where $N$ is the normal force exerted by the tube on the ball. According to Newton's second law $F_{net}=\frac{mv^{2}}{r}$, and since the normal force $N$ is what the scale reads (because of Newton's third law, the force on the scale is equal to the normal force exerted by the ball on the tube), we substitute $v^{2}$ into the centripetal - force equation: $N - mg=\frac{m\cdot2g(h - r)}{r}$ $N=mg+\frac{2mg(h - r)}{r}=mg\left(1 + \frac{2(h - r)}{r}\right)=mg\left(\frac{r+2h - 2r}{r}\right)=mg\left(\frac{2h - r}{r}\right)$

Answer:

$mg\left(\frac{2h - r}{r}\right)$