Question

Review: As a prank, someone has balanced a pumpkin at the highest point of a grain silo. The silo is topped with a hemispherical cap that is frictionless when wet. The line from the center of curvature of the cap to the pumpkin makes an angle $\mathbf{\theta }\mathbf{=}{\mathbf{0}}^{\mathbf{o}}$ with the vertical. While you happen to be standing nearby in the middle of a rainy night, a breath of wind makes the pumpkin start sliding downward from rest. It loses contact with the cap when the line from the center of the hemisphere to the pumpkin makes a certain angle with the vertical. What is this angle?

Short answer

The pumpkin will lose contact with the surface when the line joining the center of curvature with the pumpkin makes an angle of ${48.2}^{\mathrm{o}}$ with the vertical.

Step-by-step solution

Step 1:

The law of conservation of energy states that the sum of the kinetic and the mechanical energies of a system remain constant for an isolated system.

$\mathrm{\Delta K}+\mathrm{\Delta U}=0$

Here:

$\mathrm{m}=$Mass of the body

$\mathrm{a}=$Acceleration of the body

$\mathrm{\Delta U}=$Change in Potential energy

$\mathrm{\Delta K}=$Change in kinetic energy

${\mathrm{\Delta E}}_{\mathrm{int}}=$ Change in internal energy

Step 2

Let m be the mass of pumpkin and R be the radius of silo top.

The equation of motion for the pumpkin can be written as:

$\begin{array}{c}\sum \mathrm{F}={\mathrm{ma}}_{\mathrm{r}}\\ \mathrm{n}-\mathrm{mgcos\theta }=-\mathrm{m}\frac{{\mathrm{v}}^2}{\mathrm{R}}\end{array}$

When the pumpkin first loses contact with the surface, $\mathrm{n}$should be zero.

Thus, at the point where it leaves the surface, we have:

${\mathrm{v}}^2=\mathrm{Rgcos\theta }\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

Let ${\mathrm{U}}_{\mathrm{g}}=0$ in the $\mathrm{\theta }={90}^{\mathrm{o}}$ plane. Then, by applying the law of conservation of energy for the pumpkin-Earth system between the starting point and the point where the pumpkin leaves the surface, we get:

$\begin{array}{c}\mathrm{\Delta K}+\mathrm{\Delta U}=0\\ ({\mathrm{K}}_{\mathrm{f}}+{\mathrm{U}}_{\mathrm{gf}})={\mathrm{K}}_{\mathrm{i}}+{\mathrm{U}}_{\mathrm{gi}}\\ \frac{1}{2}{\mathrm{mv}}^2+\mathrm{mgRcos\theta }=0+\mathrm{mgR}\end{array}$

Here, the subscript i denotes the values at the point when $\mathrm{\theta }=0$ and the subscript frepresents the values when the pumpkin loses contact. Also, $\mathrm{K}$is the kinetic energy of the pumpkin and $\mathrm{U}$is the potential energy of the pumpkin.

Using equation (1), this becomes:

$\frac{1}{2}\mathrm{mRgcos\theta }+\mathrm{mgRcos\theta }=\mathrm{mgR}$

This reduces to:

$\begin{array}{c}\frac{3}{2}\mathrm{mRgcos\theta }=\mathrm{mgR}\\ \mathrm{cos\theta }=\frac{2}{3}\\ \mathrm{\theta }={\cos }^{-1}\frac{2}{3}\\ ={48.2}^{\mathrm{o}}.\end{array}$

The angle at which the pumpkin will lose contact with the surface is $48.2°.$