Mass Moment of Inertia

Example 1

Mass radius of gyration — solve for mass moment of inertia — Mass Moment of Inertia

A flywheel's mass radius of gyration is used to find its mass moment of inertia. Given mass (m) = 0.6000 slug; radius of gyration (k) = 2.1500 ft, determine the mass moment of inertia (I) in slug-ft^2.

Given

$mass (m) = 0.6000 slug$
$radius of gyration (k) = 2.1500 ft$

Find

mass moment of inertia (I), in slug-ft^2

Start with the thinking

The governing relation printed in this handbook section is Mass radius of gyration.
Everything except I is given, so isolate I symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
The mass radius of gyration relates a rigid body's mass moment of inertia to an equivalent concentrated mass location.

Figure 1 — schematic for Mass radius of gyration — solve for mass moment of inertia — Mass Moment of Inertia

Step-by-step solution

Step 1 — State the governing relation:
$I = m k^2$
Step 2 — Rearrange symbolically for I:
$I = m k^2$
Step 3
$List the givens: mass (m) = 0.6000 slug, radius of gyration (k) = 2.1500 ft$
Step 4 — Substitute the given values:
$I = 0.6000 2.1500^2$
Step 5 — Evaluate:
$I = 2.7735\ \text{slug-ft^2}$
Step 6 — Check: returning I = 2.7735 slug-ft^2 to
$I = m k^2$
reproduces the given quantities, and both sides carry the same units.

Answer:

I = 2.7735\ \text{slug-ft^2}

Why the other options are there

5.5470 — kept a factor of two that cancels in the correct rearrangement.
1.3868 — dropped that same factor in the other direction.
3.0509 — rounded an intermediate value before the final step.

Reference: FE Handbook — Dynamics: Mass Radius of Gyration

Example 2

Mass radius of gyration — solve for mass — Mass Moment of Inertia (2)

A rotating disk's mass radius of gyration is computed from test data. Given radius of gyration (k) = 1.7500 ft; mass moment of inertia (I) = 65.5000 slug-ft^2, determine the mass (m) in slug.

Given

$radius of gyration (k) = 1.7500 ft$
$mass moment of inertia (I) = 65.5000 slug-ft^2$

Find

mass (m), in slug

Start with the thinking

The governing relation printed in this handbook section is Mass radius of gyration.
Everything except m is given, so isolate m symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
The mass radius of gyration relates a rigid body's mass moment of inertia to an equivalent concentrated mass location.

Figure 2 — schematic for Mass radius of gyration — solve for mass — Mass Moment of Inertia (2)

Step-by-step solution

Step 1 — State the governing relation:
$I = m k^2$
Step 2 — Rearrange symbolically for m:
$m = \dfrac{I}{k^2}$
Step 3 — List the givens: radius of gyration (k) = 1.7500 ft, mass moment of inertia (I) = 65.5000 slug-ft^2.
Step 4 — Substitute the given values:
$m = \dfrac{65.5000}{1.7500^2}$
Step 5 — Evaluate:
$m = 21.3878\ \text{slug}$
Step 6 — Check: returning m = 21.3878 slug to
$I = m k^2$
reproduces the given quantities, and both sides carry the same units.

Answer:

m = 21.3878\ \text{slug}

Why the other options are there

42.7755 — kept a factor of two that cancels in the correct rearrangement.
10.6939 — dropped that same factor in the other direction.
23.5265 — rounded an intermediate value before the final step.

Reference: FE Handbook — Dynamics: Mass Radius of Gyration

Example 3

Mass radius of gyration — solve for radius of gyration — Mass Moment of Inertia (3)

A vehicle wheel's mass radius of gyration is estimated for a dynamics problem. Given mass (m) = 21.1000 slug; mass moment of inertia (I) = 183.5 slug-ft^2, determine the radius of gyration (k) in ft.

Given

$mass (m) = 21.1000 slug$
$mass moment of inertia (I) = 183.5 slug-ft^2$

Find

radius of gyration (k), in ft

Start with the thinking

The governing relation printed in this handbook section is Mass radius of gyration.
Everything except k is given, so isolate k symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
The mass radius of gyration relates a rigid body's mass moment of inertia to an equivalent concentrated mass location.

Figure 3 — schematic for Mass radius of gyration — solve for radius of gyration — Mass Moment of Inertia (3)

Step-by-step solution

Step 1 — State the governing relation:
$I = m k^2$
Step 2 — Rearrange symbolically for k:
$k = \sqrt{\dfrac{I}{m}}$
Step 3
$List the givens: mass (m) = 21.1000 slug, mass moment of inertia (I) = 183.5 slug-ft^2$
Step 4 — Substitute the given values:
$k = \sqrt{\dfrac{183.5}{21.1000}}$
Step 5 — Evaluate:
$k = 2.9490\ \text{ft}$
Step 6 — Check: returning k = 2.9490 ft to
$I = m k^2$
reproduces the given quantities, and both sides carry the same units.

Answer:

k = 2.9490\ \text{ft}

Why the other options are there

5.8980 — kept a factor of two that cancels in the correct rearrangement.
1.4745 — dropped that same factor in the other direction.
3.2439 — rounded an intermediate value before the final step.

Reference: FE Handbook — Dynamics: Mass Radius of Gyration

Example 4

Mass radius of gyration — solve for mass moment of inertia (case 2) — Mass Moment of Inertia (4)

A flywheel's mass radius of gyration is used to find its mass moment of inertia. Given mass (m) = 22.2000 slug; radius of gyration (k) = 1.8500 ft, determine the mass moment of inertia (I) in slug-ft^2.

Given

$mass (m) = 22.2000 slug$
$radius of gyration (k) = 1.8500 ft$

Find

mass moment of inertia (I), in slug-ft^2

Start with the thinking

The governing relation printed in this handbook section is Mass radius of gyration.
Everything except I is given, so isolate I symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
The mass radius of gyration relates a rigid body's mass moment of inertia to an equivalent concentrated mass location.

Figure 4 — schematic for Mass radius of gyration — solve for mass moment of inertia (case 2) — Mass Moment of Inertia (4)

Step-by-step solution

Step 1 — State the governing relation:
$I = m k^2$
Step 2 — Rearrange symbolically for I:
$I = m k^2$
Step 3
$List the givens: mass (m) = 22.2000 slug, radius of gyration (k) = 1.8500 ft$
Step 4 — Substitute the given values:
$I = 22.2000 1.8500^2$
Step 5 — Evaluate:
$I = 75.9795\ \text{slug-ft^2}$
Step 6 — Check: returning I = 75.9795 slug-ft^2 to
$I = m k^2$
reproduces the given quantities, and both sides carry the same units.

Answer:

I = 75.9795\ \text{slug-ft^2}

Why the other options are there

152.0 — kept a factor of two that cancels in the correct rearrangement.
37.9898 — dropped that same factor in the other direction.
83.5775 — rounded an intermediate value before the final step.

Reference: FE Handbook — Dynamics: Mass Radius of Gyration

Example 5

Mass radius of gyration — solve for mass (case 2) — Mass Moment of Inertia (5)

A rotating disk's mass radius of gyration is computed from test data. Given radius of gyration (k) = 0.8500 ft; mass moment of inertia (I) = 164.8 slug-ft^2, determine the mass (m) in slug.

Given

$radius of gyration (k) = 0.8500 ft$
$mass moment of inertia (I) = 164.8 slug-ft^2$

Find

mass (m), in slug

Start with the thinking

The governing relation printed in this handbook section is Mass radius of gyration.
Everything except m is given, so isolate m symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
The mass radius of gyration relates a rigid body's mass moment of inertia to an equivalent concentrated mass location.

Figure 5 — schematic for Mass radius of gyration — solve for mass (case 2) — Mass Moment of Inertia (5)

Step-by-step solution

Step 1 — State the governing relation:
$I = m k^2$
Step 2 — Rearrange symbolically for m:
$m = \dfrac{I}{k^2}$
Step 3 — List the givens: radius of gyration (k) = 0.8500 ft, mass moment of inertia (I) = 164.8 slug-ft^2.
Step 4 — Substitute the given values:
$m = \dfrac{164.8}{0.8500^2}$
Step 5 — Evaluate:
$m = 228.1\ \text{slug}$
Step 6 — Check: returning m = 228.1 slug to
$I = m k^2$
reproduces the given quantities, and both sides carry the same units.

Answer:

m = 228.1\ \text{slug}

Why the other options are there

456.2 — kept a factor of two that cancels in the correct rearrangement.
114.0 — dropped that same factor in the other direction.
250.9 — rounded an intermediate value before the final step.

Reference: FE Handbook — Dynamics: Mass Radius of Gyration

Example 6

Mass radius of gyration — solve for radius of gyration (case 2) — Mass Moment of Inertia (6)

A vehicle wheel's mass radius of gyration is estimated for a dynamics problem. Given mass (m) = 20.9000 slug; mass moment of inertia (I) = 57.3000 slug-ft^2, determine the radius of gyration (k) in ft.

Given

$mass (m) = 20.9000 slug$
$mass moment of inertia (I) = 57.3000 slug-ft^2$

Find

radius of gyration (k), in ft

Start with the thinking

The governing relation printed in this handbook section is Mass radius of gyration.
Everything except k is given, so isolate k symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
The mass radius of gyration relates a rigid body's mass moment of inertia to an equivalent concentrated mass location.

Figure 6 — schematic for Mass radius of gyration — solve for radius of gyration (case 2) — Mass Moment of Inertia (6)

Step-by-step solution

Step 1 — State the governing relation:
$I = m k^2$
Step 2 — Rearrange symbolically for k:
$k = \sqrt{\dfrac{I}{m}}$
Step 3
$List the givens: mass (m) = 20.9000 slug, mass moment of inertia (I) = 57.3000 slug-ft^2$
Step 4 — Substitute the given values:
$k = \sqrt{\dfrac{57.3000}{20.9000}}$
Step 5 — Evaluate:
$k = 1.6558\ \text{ft}$
Step 6 — Check: returning k = 1.6558 ft to
$I = m k^2$
reproduces the given quantities, and both sides carry the same units.

Answer:

k = 1.6558\ \text{ft}

Why the other options are there

3.3116 — kept a factor of two that cancels in the correct rearrangement.
0.8279 — dropped that same factor in the other direction.
1.8214 — rounded an intermediate value before the final step.

Reference: FE Handbook — Dynamics: Mass Radius of Gyration

Example 7

Mass radius of gyration — solve for mass moment of inertia (case 3) — Mass Moment of Inertia (7)

A flywheel's mass radius of gyration is used to find its mass moment of inertia. Given mass (m) = 27.3000 slug; radius of gyration (k) = 0.6500 ft, determine the mass moment of inertia (I) in slug-ft^2.

Given

$mass (m) = 27.3000 slug$
$radius of gyration (k) = 0.6500 ft$

Find

mass moment of inertia (I), in slug-ft^2

Start with the thinking

The governing relation printed in this handbook section is Mass radius of gyration.
Everything except I is given, so isolate I symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
The mass radius of gyration relates a rigid body's mass moment of inertia to an equivalent concentrated mass location.

Figure 7 — schematic for Mass radius of gyration — solve for mass moment of inertia (case 3) — Mass Moment of Inertia (7)

Step-by-step solution

Step 1 — State the governing relation:
$I = m k^2$
Step 2 — Rearrange symbolically for I:
$I = m k^2$
Step 3
$List the givens: mass (m) = 27.3000 slug, radius of gyration (k) = 0.6500 ft$
Step 4 — Substitute the given values:
$I = 27.3000 0.6500^2$
Step 5 — Evaluate:
$I = 11.5343\ \text{slug-ft^2}$
Step 6 — Check: returning I = 11.5343 slug-ft^2 to
$I = m k^2$
reproduces the given quantities, and both sides carry the same units.

Answer:

I = 11.5343\ \text{slug-ft^2}

Why the other options are there

23.0685 — kept a factor of two that cancels in the correct rearrangement.
5.7671 — dropped that same factor in the other direction.
12.6877 — rounded an intermediate value before the final step.

Reference: FE Handbook — Dynamics: Mass Radius of Gyration

Example 8

Mass radius of gyration — solve for mass (case 3) — Mass Moment of Inertia (8)

A rotating disk's mass radius of gyration is computed from test data. Given radius of gyration (k) = 0.2500 ft; mass moment of inertia (I) = 177.8 slug-ft^2, determine the mass (m) in slug.

Given

$radius of gyration (k) = 0.2500 ft$
$mass moment of inertia (I) = 177.8 slug-ft^2$

Find

mass (m), in slug

Start with the thinking

The governing relation printed in this handbook section is Mass radius of gyration.
Everything except m is given, so isolate m symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
The mass radius of gyration relates a rigid body's mass moment of inertia to an equivalent concentrated mass location.

Figure 8 — schematic for Mass radius of gyration — solve for mass (case 3) — Mass Moment of Inertia (8)

Step-by-step solution

Step 1 — State the governing relation:
$I = m k^2$
Step 2 — Rearrange symbolically for m:
$m = \dfrac{I}{k^2}$
Step 3 — List the givens: radius of gyration (k) = 0.2500 ft, mass moment of inertia (I) = 177.8 slug-ft^2.
Step 4 — Substitute the given values:
$m = \dfrac{177.8}{0.2500^2}$
Step 5 — Evaluate:
$m = 2845\ \text{slug}$
Step 6 — Check: returning m = 2,845 slug to
$I = m k^2$
reproduces the given quantities, and both sides carry the same units.

Answer:

m = 2845\ \text{slug}

Why the other options are there

5,690 — kept a factor of two that cancels in the correct rearrangement.
1,422 — dropped that same factor in the other direction.
3,129 — rounded an intermediate value before the final step.

Reference: FE Handbook — Dynamics: Mass Radius of Gyration

Example 9

Mass radius of gyration — solve for radius of gyration (case 3) — Mass Moment of Inertia (9)

A vehicle wheel's mass radius of gyration is estimated for a dynamics problem. Given mass (m) = 13.6000 slug; mass moment of inertia (I) = 79.0000 slug-ft^2, determine the radius of gyration (k) in ft.

Given

$mass (m) = 13.6000 slug$
$mass moment of inertia (I) = 79.0000 slug-ft^2$

Find

radius of gyration (k), in ft

Start with the thinking

The governing relation printed in this handbook section is Mass radius of gyration.
Everything except k is given, so isolate k symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
The mass radius of gyration relates a rigid body's mass moment of inertia to an equivalent concentrated mass location.

Figure 9 — schematic for Mass radius of gyration — solve for radius of gyration (case 3) — Mass Moment of Inertia (9)

Step-by-step solution

Step 1 — State the governing relation:
$I = m k^2$
Step 2 — Rearrange symbolically for k:
$k = \sqrt{\dfrac{I}{m}}$
Step 3
$List the givens: mass (m) = 13.6000 slug, mass moment of inertia (I) = 79.0000 slug-ft^2$
Step 4 — Substitute the given values:
$k = \sqrt{\dfrac{79.0000}{13.6000}}$
Step 5 — Evaluate:
$k = 2.4102\ \text{ft}$
Step 6 — Check: returning k = 2.4102 ft to
$I = m k^2$
reproduces the given quantities, and both sides carry the same units.

Answer:

k = 2.4102\ \text{ft}

Why the other options are there

4.8203 — kept a factor of two that cancels in the correct rearrangement.
1.2051 — dropped that same factor in the other direction.
2.6512 — rounded an intermediate value before the final step.

Reference: FE Handbook — Dynamics: Mass Radius of Gyration

Example 10

Mass radius of gyration — solve for mass moment of inertia (case 4) — Mass Moment of Inertia (10)

A flywheel's mass radius of gyration is used to find its mass moment of inertia. Given mass (m) = 24.2000 slug; radius of gyration (k) = 2.2000 ft, determine the mass moment of inertia (I) in slug-ft^2.

Given

$mass (m) = 24.2000 slug$
$radius of gyration (k) = 2.2000 ft$

Find

mass moment of inertia (I), in slug-ft^2

Start with the thinking

The governing relation printed in this handbook section is Mass radius of gyration.
Everything except I is given, so isolate I symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
The mass radius of gyration relates a rigid body's mass moment of inertia to an equivalent concentrated mass location.

Figure 10 — schematic for Mass radius of gyration — solve for mass moment of inertia (case 4) — Mass Moment of Inertia (10)

Step-by-step solution

Step 1 — State the governing relation:
$I = m k^2$
Step 2 — Rearrange symbolically for I:
$I = m k^2$
Step 3
$List the givens: mass (m) = 24.2000 slug, radius of gyration (k) = 2.2000 ft$
Step 4 — Substitute the given values:
$I = 24.2000 2.2000^2$
Step 5 — Evaluate:
$I = 117.1\ \text{slug-ft^2}$
Step 6 — Check: returning I = 117.1 slug-ft^2 to
$I = m k^2$
reproduces the given quantities, and both sides carry the same units.

Answer:

I = 117.1\ \text{slug-ft^2}

Why the other options are there

234.3 — kept a factor of two that cancels in the correct rearrangement.
58.5640 — dropped that same factor in the other direction.
128.8 — rounded an intermediate value before the final step.

Reference: FE Handbook — Dynamics: Mass Radius of Gyration

Core formulas for this FE topic

Worked exam-style examples