GRE Physics Companion: Moment of Inertia of Discrete and Continuous Bodies

GRE Physics Companion: Moment of Inertia of Discrete and Continuous Bodies

The defining equations are

(1)

for discrete masses and

(2)

for a continuous body.

1 High-value GRE facts

1. moment of inertia always refers to a specified axis.
2. For point masses, I = ∑ mr_⊥².
3. For continuous bodies, I = ∫ r_⊥² dm.
4. Mass farther from the axis matters more because of the squared radius.
5. A thin ring has I = MR² about its symmetry axis.
6. A uniform disk or solid cylinder has I = MR² about its symmetry axis.
7. A uniform rod has I = ML² about a perpendicular axis through its center.
8. A solid sphere has I = MR² about a diameter.
9. radius of gyration satisfies I = Mk_g².
10. Component moments of inertia add only when all are evaluated about the same axis.

Part I: Original GRE-style problems

Problem 1: two point masses

Two point masses m lie at distances R and 2R from an axis. Their total moment of inertia is

1. 2mR²
2. 3mR²
3. 4mR²
4. 5mR²
5. 8mR²

Problem 2: moving a point mass

A point mass initially at radius R is moved to radius 3R. Its contribution to moment of inertia changes by a factor of

1. 1∕3
2. 3
3. 6
4. 9
5. 27

Problem 3: ring versus disk

A thin ring and a uniform disk have the same mass M and radius R. About their symmetry axes,

1. I_ring = I_disk∕2
2. I_ring = I_disk
3. I_ring = 2I_disk
4. I_ring = 4I_disk
5. the result depends on angular speed

Problem 4: rod about center

A uniform thin rod of mass M and length L rotates about a perpendicular axis through its center. Its moment of inertia is

1. ML²
2. ML²
3. ML²
4. ML²
5. ML²

Problem 5: disk numerical value

A uniform disk has M = 4 kg and R = 0.50 m. Its moment of inertia about the symmetry axis is

1. 0.25 kg m²
2. 0.50 kg m²
3. 1.00 kg m²
4. 2.00 kg m²
5. 4.00 kg m²

Problem 6: radius of gyration

A body has moment of inertia I = 18 kg m² and mass M = 8 kg. Its radius of gyration is

1. 0.67 m
2. 1.0 m
3. 1.5 m
4. 2.25 m
5. 4.5 m

Problem 7: solid sphere

A uniform solid sphere of mass M and radius R has moment of inertia about a diameter

1. MR²
2. MR²
3. MR²
4. MR²
5. MR²

Problem 8: density choice

For a thin uniform plate, the most natural mass element is

1. dm = λds
2. dm = σ dA
3. dm = ρdV
4. dm = M dr
5. dm = I dA

Problem 9: scaling at fixed mass

A geometrically similar body is enlarged so every linear dimension doubles while total mass is held fixed. Its moment of inertia about the corresponding axis changes by a factor of

1. 1∕2
2. 1
3. 2
4. 4
5. 8

Problem 10: rotational kinetic energy

A body has I = 2 kg m² and rotates at ω = 5 rad∕s. Its rotational kinetic energy is

1. 5 J
2. 10 J
3. 20 J
4. 25 J
5. 50 J

Problem 11: composite rotor

A coaxial thin ring with inertia I₁ is rigidly attached to a disk with inertia I₂. The total inertia about the common axis is

1. I₁ − I₂
2. I₁I₂
3. I₁∕I₂
4. I₁ + I₂
5. 

Problem 12: same mass and center of mass

Two bodies have the same total mass and the same center of mass. Their moments of inertia about the same axis

1. must be equal
2. can be different
3. must both be zero
4. depend only on angular speed
5. depend only on torque

Part II: Complete worked solutions

Solution 1

I	= mR2 + m(2R)2	(3)
	= mR2 + 4mR2	(4)
	= 5mR2.	(5)

Answer: (D).

Solution 2

A point-mass contribution scales as r²:

(6)

Answer: (D).

Solution 3

Iring	= MR2,	(7)
Idisk	= MR2.	(8)

Therefore

(9)

Answer: (C).

Solution 4

For a uniform thin rod about its center,

(10)

Answer: (B).

Solution 5

I	= MR2	(11)
	= (4)(0.50)2	(12)
	= 0.50 kg m2.	(13)

Answer: (B).

Solution 6

kg	=	(14)
	=	(15)
	= 1.5 m.	(16)

Answer: (C).

Solution 7

A uniform solid sphere has

(17)

Answer: (B).

Solution 8

A thin plate is modeled by surface mass density:

(18)

Answer: (B).

Solution 9

For fixed mass and geometrically similar shape,

(19)

Doubling L multiplies I by four. Answer: (D).

Solution 10

Krot	= Iω2	(20)
	= (2)(25)	(21)
	= 25 J.	(22)

Answer: (D).

Solution 11

Moments of inertia about the same axis add:

(23)

Answer: (D).

Solution 12

Center of mass depends on the first spatial moment of mass, while moment of inertia depends on squared perpendicular distance. Different distributions can share the same mass and center of mass but have different I. Answer: (B).

2 GRE checklist

1. State the axis before selecting a formula.
2. Use perpendicular distance to the axis.
3. Remember the square on radius.
4. Check whether the object is a ring, disk, rod, shell, or solid sphere.
5. Use the appropriate density for continuous bodies.
6. Compare mass distributions physically before calculating.
7. Add component inertias only about a common axis.
8. Use k_g = for radius of gyration.

References

References