Question

A homogeneous body, consisting of a cylinder and a hemisphere joined at their bases, is placed with the hemispherical end on a horizontal table. $r =$ radius of hemisphere, $H =$ height of cylinder. For $\frac{r}{H} > \sqrt{x}$ the system is in stable equilibrium. Find the value of $x$.

Answer 1

2

Answer 2

Let the radius of the hemisphere be $r$ and the height of the cylinder be $H$. The body is homogeneous, so it has a uniform density $\rho$. The hemisphere and cylinder are joined at their bases of radius $r$. The hemispherical end is placed on a horizontal table.

We set up a coordinate system with the origin at the center of the base of the hemisphere (and cylinder). The z-axis is vertical, pointing upwards. The horizontal table is at $z=-r$. The hemispherical part is below the base (from $z=-r$ to $z=0$) and the cylindrical part is above the base (from $z=0$ to $z=H$).

The mass of the hemisphere is $M_h = \rho \times (\text{Volume of hemisphere}) = \rho \times \frac{2}{3}\pi r^3$.
The center of mass (CM) of a solid hemisphere of radius $r$ is located at a distance $\frac{3r}{8}$ from the center of its base along the axis of symmetry. Since our origin is at the center of the base and the hemisphere is below the base, the z-coordinate of the CM of the hemisphere is $z_h = -\frac{3r}{8}$.

The mass of the cylinder is $M_c = \rho \times (\text{Volume of cylinder}) = \rho \times \pi r^2 H$.
The CM of a solid cylinder is at the midpoint of its height along the axis of symmetry. Since the base of the cylinder is at $z=0$ and its height is $H$, the z-coordinate of the CM of the cylinder is $z_c = \frac{H}{2}$.

The z-coordinate of the CM of the combined body ($Z_{CM}$) is given by:
$Z_{CM} = \frac{M_h z_h + M_c z_c}{M_h + M_c}$
$Z_{CM} = \frac{(\rho \frac{2}{3}\pi r^3)(-\frac{3r}{8}) + (\rho \pi r^2 H)(\frac{H}{2})}{\rho \frac{2}{3}\pi r^3 + \rho \pi r^2 H}$
We can cancel out the common factor $\rho \pi r^2$:
$Z_{CM} = \frac{(\frac{2}{3}r)(-\frac{3r}{8}) + H(\frac{H}{2})}{\frac{2}{3}r + H}$
$Z_{CM} = \frac{-\frac{6r^2}{24} + \frac{H^2}{2}}{\frac{2r}{3} + H}$
$Z_{CM} = \frac{-\frac{r^2}{4} + \frac{H^2}{2}}{\frac{2r + 3H}{3}}$
$Z_{CM} = \frac{\frac{2H^2 - r^2}{4}}{\frac{2r + 3H}{3}}$
$Z_{CM} = \frac{3(2H^2 - r^2)}{4(2r + 3H)}$

The body is resting on the hemispherical end on a horizontal table. The point of contact in the upright position is the lowest point of the hemisphere, which is at $z=-r$. This point is on the spherical surface of radius $r$. The center of curvature of the hemispherical surface at this point is the center of the hemisphere, which is at $z=0$ in our coordinate system.

For a body with a curved base resting on a horizontal plane, the condition for stable equilibrium is that the center of mass of the body must be below the center of curvature of the base at the point of contact.
In this case, the center of curvature is at $z=0$. The center of mass is at $z=Z_{CM}$.
For stable equilibrium, the z-coordinate of the CM must be less than the z-coordinate of the center of curvature.
$Z_{CM} < 0$

Substituting the expression for $Z_{CM}$:
$\frac{3(2H^2 - r^2)}{4(2r + 3H)} < 0$
Since $r$ and $H$ are positive lengths, $4(2r + 3H)$ is always positive. Also, $3$ is positive.
So, the inequality reduces to:
$2H^2 - r^2 < 0$
$2H^2 < r^2$
$\frac{r^2}{H^2} > 2$
$\left(\frac{r}{H}\right)^2 > 2$
Since $r$ and $H$ are positive, we can take the square root:
$\frac{r}{H} > \sqrt{2}$

The question states that the system is in stable equilibrium for $\frac{r}{H} > \sqrt{x}$.
Comparing our condition $\frac{r}{H} > \sqrt{2}$ with the given condition $\frac{r}{H} > \sqrt{x}$, we find that $x=2$.