Question

Two identical uniform solid spheres areattached by a solid uniform thin rod, as shown in the figure. Therod lies on a line connecting the centers of mass of the twospheres. The axes A, B, C, and D are in the plane of the page(which also contains the centers of mass of the spheres and therod), while axes E and F (represented by black dots) areperpendicular to the page.

Rank the momentsof inertia of this object about the axesindicated.

Rank from largestto smallest. To rank items as equivalent, overlap them.A,B,C,D,E,F

Answer 1

Concepts and reason

The main concept of this question is moment of inertia of a uniform solid sphere.

Initially, calculate the moment of inertia of the spheres about axis F. Later, use theorem of parallel axis for the rest of the axis. Finally, rank them based on moment of inertia from largest to smallest.

Fundamentals

Consider two identical solid spheres of mass m and radius r attached by a rod of length l.

The moment of inertia ${I_1}$ of the sphere about an axis passing through the center of the sphere is,

${I_1} = \frac{2}{5}m{r^2}$

Here, m is the mass, r is the radius, and ${I_1}$ is the moment of inertia.

The moment of inertia ${I_2}$. of the rod about an axis passing through its center and perpendicular to it is,

${I_2} = \frac{1}{{12}}M{l^2}$

Here, M is the mass of rod, l is the length, and ${I_2}$ is the moment of inertia.

The theorem of parallel axis states that the moment of inertia of a rigid body about any axis is equal to the sum of the moment of inertia of a rigid body about a parallel axis passing through the center of mass and the product of mass of the object with square of the distance between two axes.

Find the moment of inertia of the sphere about axis F, E and A.

Let F be the axis passing through the center of sphere ${S_1}$ and perpendicular to the plane of page.

The moment of inertia ${I_{1F}}$ of the sphere ${S_1}$about axis F is,

${I_{1F}} = \frac{2}{5}m{r^2}$

Apply theorem of parallel axis for the moment of inertia ${I_{2F}}$ of the sphere ${S_2}$about axis F,

$\begin{array}{c}\\{I_{2F}} = \frac{2}{5}m{r^2} + m{\left( {r + l + r} \right)^2}\\\\ = \frac{2}{5}m{r^2} + m{\left( {2r + l} \right)^2}\\\end{array}$

On the same line, apply theorem of parallel axis for the moment of inertia ${I_{3F}}$of the sphere ${S_2}$about axis F,

${I_{3F}} = \frac{1}{{12}}M{l^2} + M{\left( {r + \frac{l}{2}} \right)^2}$

Total moment of inertia ${I_F}$ about axis F is,

$\begin{array}{c}\\{I_F} = {I_{1F}} + {I_{2F}} + {I_{3F}}\\\\ = \frac{2}{5}m{r^2} + \left\{ {\frac{1}{{12}}M{l^2} + M{{\left( {r + \frac{l}{2}} \right)}^2}} \right\} + \left\{ {\frac{2}{5}m{r^2} + m{{\left( {2r + l} \right)}^2}} \right\}\\\\ = \frac{4}{5}m{r^2} + m{\left( {2r + l} \right)^2} + \frac{1}{{12}}M{l^2} + M{\left( {r + \frac{l}{2}} \right)^2}\\\end{array}$

Due to symmetry, the moment of inertia of the first sphere about a central axis in the plane of paper and parallel to axis A is same as that of axis F.

So, total moment of inertia ${I_A}$ about axis A is,

$\begin{array}{c}\\{I_A} = {I_{1A}} + {I_{2A}} + {I_{3A}}\\\\ = \left\{ {\frac{2}{5}m{r^2} + m{{\left( {r + \frac{l}{2}} \right)}^2}} \right\} + \frac{1}{{12}}M{l^2} + \left\{ {\frac{2}{5}m{r^2} + m{{\left( {r + \frac{l}{2}} \right)}^2}} \right\}\\\\ = \frac{4}{5}m{r^2} + 2m{\left( {r + \frac{l}{2}} \right)^2} + \frac{1}{{12}}M{l^2}\\\\ = \frac{4}{5}m{r^2} + m{\left( {2r + l} \right)^2} + \frac{1}{{12}}M{l^2}\\\end{array}$

Here ${I_{1A}}$, ${I_{2A}}$and ${I_{3A}}$ are moment of inertia of first, second sphere and rod about axis A.

The moment of inertia ${I_B}$ of the sphere about axis B is,

$\begin{array}{c}\\{I_B} = {I_{1B}} + {I_{2B}} + {I_{3B}}\\\\ = \left\{ {\frac{2}{5}m{r^2} + m{{\left( {r + l} \right)}^2}} \right\} + \frac{1}{{12}}M{l^2} + M{\left( {\frac{l}{2}} \right)^2} + \left\{ {\frac{2}{5}m{r^2} + m{r^2}} \right\}\\\\ = \frac{4}{5}m{r^2} + m{\left( {r + l} \right)^2} + \frac{1}{{12}}M{l^2} + M\frac{{{l^2}}}{4} + m{r^2}\\\end{array}$

The axes A and E are the central axes of the rod. Due to symmetry, the moment of inertia of the system about axes A and E is same.

On the same line, the moment of inertia of the system about axes C and F is same.

The axis D has lowest moment of inertia, because the perpendicular distance from the axis of rotation never exceeds the radius of the one of the balls.

From the order from highest to lowest is,

${\rm{C}} = {\rm{F}} > {\rm{B}} > {\rm{A}} = {\rm{E}} > {\rm{D}}$.

Ans:

The order of axis from largest to smallest is ${\rm{C}} = {\rm{F}} > {\rm{B}} > {\rm{A}} = {\rm{E}} > {\rm{D}}$ .