GRE Physics Companion: Parallel-Axis and Perpendicular-Axis Theorems

GRE Physics Companion: Parallel-Axis and Perpendicular-Axis Theorems

The two central relations are

(1)

for a shifted parallel axis, and

(2)

for a planar lamina with mutually perpendicular axes through the same point.

1 High-value GRE facts

1. The parallel-axis theorem requires parallel axes.
2. The known reference axis must pass through the center of mass.
3. d is the perpendicular distance between the two axes.
4. Shifting away from the center of mass always increases I by Md².
5. The perpendicular-axis theorem applies only to planar laminas.
6. The three perpendicular-axis theorem axes must intersect at the same point.
7. For a thin disk, I_z = MR² and I _diameter = MR².
8. For a thin ring, I_z = MR² and I _diameter = MR².
9. A disk tangent axis perpendicular to the plane has I = MR².
10. A disk tangent axis in the plane has I = MR².

Part I: Original GRE-style problems

Problem 1: shifted axis

A body has I_CM = 2I₀ about an axis through its center of mass. A parallel axis is displaced by d, and Md² = 3I ₀. The shifted-axis moment is

1. I₀
2. 2I₀
3. 3I₀
4. 5I₀
5. 6I₀

Problem 2: minimum parallel-axis inertia

Among all axes parallel to a chosen direction, the smallest moment of inertia occurs for the axis

1. farthest from the body
2. through the center of mass
3. tangent to the body
4. through the largest mass element
5. with maximum angular speed

Problem 3: rod about an end

A uniform thin rod of mass M and length L has moment of inertia about a perpendicular axis through one end equal to

1. ML²
2. ML²
3. ML²
4. ML²
5. ML²

Problem 4: disk tangent normal axis

A uniform disk of mass M and radius R rotates about an axis perpendicular to the disk and tangent to its rim. Its moment of inertia is

1. MR²
2. MR²
3. MR²
4. MR²
5. MR²

Problem 5: disk diameter

A uniform thin disk has moment of inertia about any diameter equal to

1. MR²
2. MR²
3. MR²
4. MR²
5. MR²

Problem 6: ring diameter

A thin ring has moment of inertia about any diameter equal to

1. MR²
2. MR²
3. MR²
4. MR²
5. 2MR²

Problem 7: applicability

Which body permits direct use of the perpendicular-axis theorem?

1. a uniform solid sphere
2. a thick solid cylinder about arbitrary axes
3. a thin rectangular plate
4. a three-dimensional irregular rock
5. a solid cone

Problem 8: rectangular plate

A thin rectangular plate has central in-plane moments I_x = 2I₀ and I_y = 5I₀. Its perpendicular central moment is

1. 3I₀
2. 5I₀
3. 7I₀
4. 10I₀
5. 25I₀

Problem 9: disk tangent in-plane axis

A thin disk rotates about a tangent line lying in its plane. Its moment of inertia is

1. MR²
2. MR²
3. MR²
4. MR²
5. MR²

Problem 10: inverse shift

A body has I = 10 kg m² about a shifted axis. Its mass is M = 8 kg and the parallel-axis separation is d = 0.50 m. The center of mass moment is

1. 2 kg m²
2. 4 kg m²
3. 6 kg m²
4. 8 kg m²
5. 12 kg m²

Problem 11: radius of gyration

If k_g,CM = 0.30 m and a parallel axis is displaced by d = 0.40 m, the new radius of gyration is

1. 0.10 m
2. 0.35 m
3. 0.50 m
4. 0.70 m
5. 1.00 m

Problem 12: wrong theorem

A student tries to use I_z = I_x + I_y for a solid sphere. The main error is that

1. a sphere has no center of mass
2. the sphere is not a planar lamina
3. the axes are not perpendicular
4. the sphere has no moment of inertia
5. the theorem requires zero mass

Part II: Complete worked solutions

Solution 1

The parallel-axis theorem gives

(3)

Answer: (D).

Solution 2

Since

(4)

and Md² ≥ 0, the minimum occurs at d = 0, which is the center of mass axis. Answer: (B).

Solution 3

Iend	= ML2 + M2	(5)
	= ML2.	(6)

Answer: (D).

Solution 4

For the central perpendicular axis,

(7)

Shift by R:

(8)

Answer: (E).

Solution 5

For a thin disk,

(9)

Symmetry gives I_x = I_y, so

(10)

Answer: (B).

Solution 6

For a thin ring,

(11)

By symmetry,

(12)

Answer: (B).

Solution 7

The perpendicular-axis theorem requires a planar lamina. A thin rectangular plate satisfies that condition. Answer: (C).

Solution 8

(13)

Answer: (C).

Solution 9

First use the perpendicular-axis theorem to obtain the central diameter moment:

(14)

Then shift by R:

(15)

Answer: (D).

Solution 10

ICM	= I − Md2	(16)
	= 10 − 8(0.50)2	(17)
	= 10 − 2	(18)
	= 8 kg m2.	(19)

Answer: (D).

Solution 11

Use

(20)

Then

kg	=	(21)
	= 0.50 m.	(22)

Answer: (C).

Solution 12

The perpendicular-axis theorem relies on all mass lying in one plane. A solid sphere is three-dimensional. Answer: (B).

2 GRE checklist

1. Parallel-axis theorem: axes parallel, known axis through the center of mass.
2. Use the perpendicular distance between axes for d.
3. Shift away from the center of mass by adding Md².
4. Perpendicular-axis theorem: planar lamina only.
5. All three perpendicular-axis theorem axes must meet at the same point.
6. Use symmetry only when the geometry actually supports it.
7. A tangent axis perpendicular to a disk and a tangent axis in its plane are different problems.
8. For composite bodies, shift every component to the final common axis before adding or subtracting.

References

References