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Theorems on moment of inertia (Application)

Module by: Sunil Kumar Singh. E-mail the author

Summary: Solving problems is an essential part of the understanding process.

Questions and their answers are presented here in the module text format as if it were an extension of the theoretical treatment of the topic. The idea is to provide a verbose explanation of the solution, detailing the application of theory. Solution presented here, therefore, is treated as the part of the understanding process – not merely a Q/A session. The emphasis is to enforce ideas and concepts, which can not be completely absorbed unless they are put to real time situation.

Representative problems and their solutions

We discuss problems, which highlight certain aspects of the application of theorem on moment of inertia. For this reason, questions are categorized in terms of the characterizing features pertaining to the questions :

  • Theorems on moment of inertia
  • Nature of MI
  • Part of a rigid body
  • Geometric shapes formed from wire

Theorems on moment of inertia

Example 1

Problem : A uniform circular disk of radius "R" lies in "xy" - plane such that its center is at the origin of coordinate system as shown in the figure. Its MI about the axis defined as x = R and y = 0 is equal to its MI about axis defined as y = d and z = 0. Then, find “d”.

Figure 1: The circular disk lies in “xy” - plane.
MI of uniform circular disk
 MI of uniform circular disk  (tmi1.gif)

Solution : First, we need to visualize the two axes as defined. Then, we find disk's MI about them. Finally, we equate the MIs as given in the question to find "d".

The coordinates x = R and y = 0 define the axis in "xz" - plane as "y" coordinate is zero. Further, the axis intersects x-axis at x = R = a constant. This axis (z'), therefore, is parallel to z-axis as shown in the figure.

Figure 2: The location of first axis is on x-axis.
MI of uniform circular disk
 MI of uniform circular disk  (tmi2.gif)

Note that z axis passes through center of mass. Thus by applying theorem of parallel axes, the MI of the disk about z' by the theorem of parallel axes is :

I z’ = I z + M R 2 I z’ = I z + M R 2

I z’ = M R 2 2 + M R 2 = 3 M R 2 2 I z’ = M R 2 2 + M R 2 = 3 M R 2 2

The coordinates y = d and z = 0 define the axis in "xy" - plane as "z" coordinate is zero. Further, this axis intersects y-axis at y = d = a constant. This axis (x'), therefore, is parallel to x-axis as shown in the figure.

Figure 3: The location of second axis is on y-axis.
MI of uniform circular disk
 MI of uniform circular disk  (tmi3a.gif)

The MI of the disk about x' by the theorem of parallel axes :

I x’ = I x + M R 2 I x’ = I x + M R 2

I x’ = M R 2 4 + M d 2 I x’ = M R 2 4 + M d 2

According to the question,

I x’ = I z’ I x’ = I z’

M R 2 4 + M d 2 = 3 M R 2 2 M R 2 4 + M d 2 = 3 M R 2 2

M d 2 = 3 M R 2 2 - M R 2 4 = 5 M R 2 4 M d 2 = 3 M R 2 2 - M R 2 4 = 5 M R 2 4

d = 5 R 2 d = 5 R 2

Example 2

Problem : Two identical uniform rods, each of mass "M" and length "L", are joined together to form a cross. Find the MI of the cross about the angle bisector "AA'" as shown in the figure.

Figure 4: The MI of a cross of two rods at right angle.
MI of a cross
 MI of a cross  (tmi5.gif)

Solution : MI of each of the rod about perpendicular axis, which passes through COM (O) is :

I rod = M L 2 12 I rod = M L 2 12

The MI of two rods joined together as a cross about perpendicular axis through the common point is :

I cross = 2 I rod = 2 x M L 2 12 = M L 2 6 I cross = 2 I rod = 2 x M L 2 12 = M L 2 6

The cross can be treated as a planar object. As such, we can apply theorem of perpendicular axes to find MI about the bisector of the right angle. We consider another axis BB' perpendicular to AA' in the plane of cross. Then, according to theorem of perpendicular axes,

Figure 5: The MI of a cross of two rods at right angle.
MI of a cross about angle bisector
 MI of a cross about angle bisector  (tmi4.gif)

I cross = I AA’ + I BB’ I cross = I AA’ + I BB’

The MIs about the two bisectors are equal by the symmetry of the cross about them. Hence,

I AA’ = I BB’ I AA’ = I BB’

Thus,

I cross = I AA’ + I BB’ = 2 I AA’ I cross = I AA’ + I BB’ = 2 I AA’

I AA’ = I cross 2 = M L 2 12 I AA’ = I cross 2 = M L 2 12

Nature of MI

Example 3

Problem : The MI of a solid sphere is calculated about an axis parallel to one of its diameter, at a distance "x" from the origin. The values are plotted against distance. Which of the four plots, as shown below, correctly describes the variation of MI with respect to "x" for the solid sphere.

Figure 6: The MI of a sphere about an axis parallel to central axis.
Plots of MI .vs. distance
 Plots of MI .vs. distance  (tmi6.gif)

Solution : It is evident that farther the axis from the origin, greater is MI about that axis. We can, thus, conclude that the plot between MI and the distance should be increasing one with increasing distance ("x"). Thus, options (a) and (d) are incorrect. Now, the MI of solid sphere about an axis parallel to one of diameters, at a distance "x" from the origin is given as :

Figure 7: The MI of a sphere about an axis parallel to central axis.
Plots of MI .vs. distance
 Plots of MI .vs. distance  (tmi7.gif)

I = I O + M x 2 I = I O + M x 2

where I O I O is MI about one of the diameters and is a constant for the given solid sphere. Thus, the equation above takes the form of the equation of parabola,

y = m x 2 + c y = m x 2 + c

The correct plot, therefore, is an increasing parabola as shown in Figure (c).

Part of a rigid body

Example 4

Problem : Four holes of radius "L/4" are made in a thin square plate of mass "M" and side "L" in "xy" - plane, as shown in the figure. Find its MI about z – axis.

Figure 8: Four holes of radius "L/4" are made in a thin square plate of mass "M" and side "L".
MI of remaining body
 MI of remaining body  (tmi8.gif)

Solution : MI is a scalar quantity. It means that MI has no directional attribute. Thus, we can conclude that MI of the complete plate is equal to the sum of MIs of remaining plate and that of the four circular disks taken out from the plate about z-axis. Hence, MI of remaining plate is obtained as :

I (MI of remaining plate) = Iz (MI of complete plate) - 4 x Icz (MI of circular disks) I (MI of remaining plate) = Iz (MI of complete plate) - 4 x Icz (MI of circular disks)

Now, according to theorem of perpendicular axes, the MI of complete square about z-axis is two times its MI about x-axis :

I z = 2 x M L 2 12 = M L 2 6 I z = 2 x M L 2 12 = M L 2 6

Further, according to theorem of parallel axes, the MI of circular disk of mass “m” and radius “L/4” about the same z-axis is :

Figure 9: Four holes of radius "L/4" are made in a thin square plate of mass "M" and side "L".
MI of remaining body
 MI of remaining body  (tmi9.gif)

I cz = m ( L 4 ) 2 + m ( 2 L 4 ) 2 I cz = m ( L 4 ) 2 + m ( 2 L 4 ) 2

I cz = m L 2 16 + 2 m L 2 16 = 3 m L 2 16 I cz = m L 2 16 + 2 m L 2 16 = 3 m L 2 16

The mass of the disk "m" is given by multiplying areal density with the area of the circular disk,

m = M L 2 x π ( L 4 ) 2 = π M 16 m = M L 2 x π ( L 4 ) 2 = π M 16

Substituting for "m", we have :

I cz = 3 m L 2 16 = 3 π M L 2 16 x 16 I cz = 3 m L 2 16 = 3 π M L 2 16 x 16

Thus, the required MI is :

I = M L 2 6 - 4 x 3 π M L 2 16 x 16 = 16 x 16 M L 2 - 12 x 6 π M L 2 6 x 16 x 16 I = M L 2 6 - 4 x 3 π M L 2 16 x 16 = 16 x 16 M L 2 - 12 x 6 π M L 2 6 x 16 x 16

I = 32 M L 2 - 9 π M L 2 192 I = 32 M L 2 - 9 π M L 2 192

Geometric shapes formed from wire

Example 5

Problem : A thin wire of length "L" and uniform linear mass density "λ" is bent into a circular loop. Find the MI of the loop about an axis AA' as shown in the figure.

Figure 10: MI of a circular loop about a tangent.
MI of a circular loop
 MI of a circular loop  (tmi10.gif)

Solution : The MI of the circular loop about a tangential axis AA' is found out by the theorem of parallel axis :

Figure 11: MI of a circular loop about a tangent.
MI of a circular loop
 MI of a circular loop  (tmi11.gif)

I AA’ = I XX’ + M R 2 I AA’ = I XX’ + M R 2

where IXX' is the MI about an axis which is one of the diameters of the ring and is parallel to tangential axis AA'. Now, MI about the diameter can be calculated applying theorem of perpendicular axes :

I XX’ = I O 2 I XX’ = I O 2

where I O I O is MI about an axis perpendicular to the surface of loop and passing through center of mass (ZZ' axis). It is given by the expression :

Figure 12: MI of a circular loop about its perpendicular central axis.
MI of a circular loop
 MI of a circular loop  (tmi12.gif)

I O = M R 2 I O = M R 2

Combining two equations, we have :

I XX’ = M R 2 2 I XX’ = M R 2 2

Putting this in the expression of I AA' I AA' , we have :

I AA’ = M R 2 2 + M R 2 = 3 M R 2 2 I AA’ = M R 2 2 + M R 2 = 3 M R 2 2

Now, we are required to find mass and radius from the given data. Here,

M = λ L M = λ L

Since the wire is bent into a circle, the perimeter of the circle is equal to the length of the wire. Hence,

2 π R = L R = L 2 π 2 π R = L R = L 2 π

Putting these values in the expression of MI about AA' axis is :

I AA’ = 3 M R 2 2 = 3 λ L x L 2 2 ( 2 π ) 2 = 3 λ L 3 8 π 2 I AA’ = 3 M R 2 2 = 3 λ L x L 2 2 ( 2 π ) 2 = 3 λ L 3 8 π 2

Example 6

Problem : A uniform wire of mass "M" and length "L" is bent into a regular hexagonal loop. Find its MI about an axis passing through center of mass and perpendicular to its surface.

Solution : We find MI of hexagonal loop in following three steps :

  1. Treat each side as a wire of mass M/6 and length L/6 and find its MI about perpendicular axis through center of mass of the side.
  2. Apply theorem of parallel axes to find MI of the side about perpendicular axis through center of mass of the loop.
  3. Add MIs of individual sides to find the MI of the loop about perpendicular axis through center of mass.

The MI of the side about perpendicular axis through center of mass of the side is :

Figure 13: MI of a regular hexagonal loop is sum of MIs of its sides.
MI of a regular hexagonal loop
 MI of a regular hexagonal loop  (tmi14.gif)

I AA’ = M 6 x ( L 6 ) 2 12 = M L 2 12 x 6 3 I AA’ = M 6 x ( L 6 ) 2 12 = M L 2 12 x 6 3

By the geometry of a side with respect to center, the traingle OPQ is a right angle triangle. We have perpendicular distance "R" between axes AA' and ZZ' as :

Figure 14: MI of a regular hexagonal loop is sum of MIs of its sides.
MI of a side of regular hexagonal loop
 MI of a side of regular hexagonal loop  (tmi15.gif)

tan 60 ° = R L 12 tan 60 ° = R L 12

R = L 12 x 3 = 3 12 L R = L 12 x 3 = 3 12 L

Applying theorem of parallel axes, the MI about perpendicular axis through center of mass of the loop is :

I O = I zz’ = I AA’ + M 6 x R 2 I O = M L 2 12 x 6 3 + M 6 x ( 3 L 12 ) 2 I O = M L 2 12 x 6 3 + 3 M L 2 12 2 x 6 I O = I zz’ = I AA’ + M 6 x R 2 I O = M L 2 12 x 6 3 + M 6 x ( 3 L 12 ) 2 I O = M L 2 12 x 6 3 + 3 M L 2 12 2 x 6

I O = M L 2 + 9 M L 2 12 x 6 3 = 10 M L 2 12 x 6 3 I O = M L 2 + 9 M L 2 12 x 6 3 = 10 M L 2 12 x 6 3

Now, adding MIs of each side, the required MI about perpendicular axis through center of mass of the loop is :

I = 6 I O = 6 x 10 M L 2 12 x 6 3 = 5 M L 2 216 I = 6 I O = 6 x 10 M L 2 12 x 6 3 = 5 M L 2 216

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