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Two identical uniform solid spheres are attached by a solid uniform thin rod, as shown in the figure. The rod lies on a line connecting the centers of mass of the two spheres. 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 the rod), while axes E and F (represented by black dots) are perpendicular to the page.

Rank the moments of inertia of this object about the axes indicated.

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

Revelant knowledge

The moment of inertia of a rigid body, also known as mass moment of inertia, angular mass, second moment of mass, or, more precisely, rotational inertia, is a quantity that determines the torque required for a desired angular acceleration about a rotational axis, in the same way that mass determines the force required for a desired acceleration. It relies on the mass distribution of the body and the axis chosen, with higher moments necessitating more torque to affect the rate of rotation.

Answer:C = F > B > E = A > DExplanation:The moment of inertia I is given by

I = mr^{2}Where: m is the mass and r is the distance of the center of mass from the axis rotation.

Since the mass is the same for all cases, the differences in the moment of inertia in the different axes will depend on the distance r from the axis of rotation.

The moment of inertia at C and F are the largest because those are the axes that are farthest from the center of mass. Next to C and F is B. Then followed by E and A. The axis where the moment of inertia is the least is at D because the distance from the axis of rotation exceeds the radius of the balls.

This question’s main concept is Moment of Inertia for a uniform solid sphere.

Calculate the moment of inertia for all spheres around axis F. Then, apply the theorem to the rest of axis. Finally, rank them according to their moment of inertia.

Consider two identical solid spheres with mass

m, radiusr, and attached to a rod ofl.The moment of inertia (equa_tag_0)

[katex]{I_1} = \frac{2}{5}m{r^2}[/katex]

Here, m is a mass, r the radius and [katex]{I_1}[/katex] are the values.

The moment of inertia [katex]{I_2}[/katex]

[katex]{I_2} = \frac{1}{{12}}M{l^2}[/katex]

Here M is the rod’s mass, l the length and [katex]{I_2}[/katex] are the parameters.

The theorem about parallel axis says that the moment inertia for a rigid object about an axis is equal the sum of its moment of inertia about an axis passing through its center of mass and the product mass of the object with squared of the distance between the axes.

Locate the moment of inertia around the axis , and

.

Let F be the axis that passes through the center of the sphere [katex]{S_1}[/katex]

The moment of inertia (equa_tag_7)

[katex]{I_{1F}} = \frac{2}{5}m{r^2}[/katex]

Theorem of parallel-axis for moment of inertia [katex]{I_{2F}}[/katex]

[katex]\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}[/katex]

Apply the theorem for parallel axis to the moment of inertia [katex]{I_{3F}}[/katex]

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

Total moment of inertia (equa_tag_13)

[katex]\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}[/katex]

Symmetry means that the moment of inertia for the first sphere around a central axis of the plane of paper parallel to A is the same as the moment for axis F.

Total moment of inertia (equa_tag_15)

[katex]\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}[/katex]

Here [katex]{I_{1A}}[/katex]

The moment of inertia [katex]{I_B}[/katex]

[katex]\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}[/katex]

The rod’s central axes are

A, and . Because of symmetry, there is no moment of inertia for the system around axesAorE.The moment of inertia for the system around axes

CandFare the same.Because the distance perpendicular to the axis from the axis rotation is never greater than the radius of one of the balls, the axis D has the lowest moment of inertia.

The order of highest to lowest is:

[katex]{\rm{C}} = {\rm{F}} > {\rm{B}} > {\rm{A}} = {\rm{E}} > {\rm{D}}[/katex]Ans:

The order in which axis are arranged from largest to most is [katex]{\rm{C}} = {\rm{F}} > {\rm{B}} > {\rm{A}} = {\rm{E}} > {\rm{D}}[/katex].