Question

A thin hemispherical shell of mass M and radius R is resting on a fixed sharp support as shown. Time period of small oscillations of the shell is:

Answer 1

2\pi \sqrt{\frac{4R}{3g}}

Answer 2

The time period of small oscillations of a physical pendulum is given by $T = 2\pi \sqrt{\frac{I}{mgd}}$, where $I$ is the moment of inertia about the pivot point, $m$ is the mass of the object, $g$ is the acceleration due to gravity, and $d$ is the distance between the pivot point and the center of mass.

For a thin hemispherical shell of mass M and radius R, the center of mass is located at a distance $R/2$ from the center of the base along the axis of symmetry. Let the support point be O, which is at the bottom of the hemisphere. In the equilibrium position, the center of mass G is vertically above O. The distance between the pivot point O and the center of mass G is $d = OG$. The center of the spherical surface C is at a distance R from O. The center of the base B is at a distance R from O. The center of mass G is at a distance R/2 from the center of the base B. Since O, B, and G are collinear in equilibrium, and B is between O and G, the distance of G from O is $OG = OB - GB = R - R/2 = R/2$. So, $d = R/2$.

We need to find the moment of inertia of the hemispherical shell about the pivot point O. We can use the parallel axis theorem: $I_O = I_G + Md^2$, where $I_G$ is the moment of inertia about an axis through the center of mass G parallel to the axis of rotation through O. The axis of rotation is horizontal. We need to find the moment of inertia of a thin hemispherical shell about an axis through its center of mass perpendicular to the axis of symmetry. The moment of inertia of a thin spherical shell of mass $2M$ and radius R about a diameter is $\frac{2}{3}(2M)R^2 = \frac{4}{3}MR^2$. The moment of inertia of a thin hemispherical shell of mass M about an axis through the center of the spherical surface C and perpendicular to the axis of symmetry is $I_C = \frac{2}{3}MR^2$. The center of mass G is at a distance $CG = R/2$ from C along the axis of symmetry. Using the parallel axis theorem, $I_C = I_G + M(CG)^2$, where $I_G$ is the moment of inertia about an axis through G parallel to the axis through C. $I_G = I_C - M(CG)^2 = \frac{2}{3}MR^2 - M(R/2)^2 = \frac{2}{3}MR^2 - \frac{1}{4}MR^2 = (\frac{8-3}{12})MR^2 = \frac{5}{12}MR^2$. This is the moment of inertia about an axis through the center of mass G perpendicular to the axis of symmetry. The axis of rotation through O is horizontal, and the axis through G parallel to this is also horizontal, which is perpendicular to the vertical axis of symmetry. So, $I_G = \frac{5}{12}MR^2$.

Now, we can find the moment of inertia about the pivot O: $I_O = I_G + Md^2 = \frac{5}{12}MR^2 + M(R/2)^2 = \frac{5}{12}MR^2 + \frac{1}{4}MR^2 = \frac{5}{12}MR^2 + \frac{3}{12}MR^2 = \frac{8}{12}MR^2 = \frac{2}{3}MR^2$.

The time period of small oscillations is: $T = 2\pi \sqrt{\frac{I_O}{Mgd}} = 2\pi \sqrt{\frac{\frac{2}{3}MR^2}{Mg(R/2)}} = 2\pi \sqrt{\frac{\frac{2}{3}R^2}{\frac{1}{2}gR}} = 2\pi \sqrt{\frac{2}{3}R^2 \cdot \frac{2}{gR}} = 2\pi \sqrt{\frac{4R}{3g}}$.

$T = 2\pi \sqrt{\frac{4R}{3g}}$.

Explanation of the solution:

1. Identify the pivot point and the center of mass.
2. Calculate the distance $d$ between the pivot and the center of mass.
3. Calculate the moment of inertia $I$ about the pivot point using the parallel axis theorem, $I_O = I_G + Md^2$.
4. Use the formula for the time period of a physical pendulum: $T = 2\pi \sqrt{\frac{I}{mgd}}$.

Answer: $2\pi \sqrt{\frac{4R}{3g}}$