GRE Physics Companion: Gyroscopes, Precession, and Nutation

GRE Physics Companion: Gyroscopes, Precession, and Nutation

The fundamental relation is

(1)

For steady precession of a fixed-magnitude angular-momentum vector,

(2)

For a rapidly spinning heavy top,

(3)

1 High-value GRE facts

1. A torque perpendicular to angular momentum primarily changes its direction.
2. Steady vector precession satisfies τ = Ω_p × L.
3. Fast gravitational precession scales inversely with spin angular momentum.
4. Faster spin usually means slower gravitational precession.
5. Spin, precession, and nutation are distinct motions.
6. Nutation is variation of the inclination angle.
7. Exact heavy-top motion generally contains both precession and nutation.
8. For the symmetric top, ϕ and ψ are cyclic coordinates.
9. Nutation turning points occur at E = U_eff(𝜃).
10. A forced precessing rotor exerts an equal and opposite gyroscopic reaction torque on its support.

Part I: Original GRE-style problems

Problem 1: perpendicular torque

A torque remains perpendicular to a constant-magnitude angular momentum vector. Its primary effect is to change

1. only the mass
2. the direction of angular momentum
3. the moment of inertia only
4. the center of mass position only
5. gravitational acceleration

Problem 2: steady precession magnitude

If the angle between Ω_p and L is 𝜃, then

1. τ = Ω_pL
2. τ = Ω_pL sin 𝜃
3. τ = Ω_p∕L
4. τ = L∕Ω_p
5. τ = Ω_pL cos ²𝜃

Problem 3: fast heavy top

For a rapidly spinning heavy top,

(4)

is approximately

1. I₃ω₃∕(Mgℓ)
2. Mgℓ∕(I₃ω₃)
3. MgℓI₃ω₃
4. g∕ω₃²
5. ω₃

Problem 4: effect of doubling spin

Within the fast-top approximation, doubling ω₃ changes Ω_p by a factor of

1. 1∕4
2. 1∕2
3. 1
4. 2
5. 4

Problem 5: gravity torque

A top has center of mass distance ℓ from its pivot and inclination 𝜃 from vertical. The gravity torque magnitude is

1. Mgℓ
2. Mgℓ cos 𝜃
3. Mgℓ sin 𝜃
4. Mg∕ℓ
5. I₃ω₃

Problem 6: nutation

Nutation refers to

1. spin about the symmetry axis
2. oscillation of the inclination angle
3. translation of the pivot
4. constant angular momentum magnitude
5. zero torque

Problem 7: cyclic spin coordinate

For the heavy symmetric top,

(5)

equals

1. I₃ω₃
2. Mgℓ
3. I₁𝜃
4. I₁ϕ
5. zero necessarily

Problem 8: precession rate from constants

For the symmetric top,

(6)

equals

1. (p_ϕ − p_ψ cos 𝜃)∕(I₁ sin ²𝜃)
2. p_ϕ∕I₃
3. p_ψ∕I₁
4. Mgℓ∕I₁
5. zero

Problem 9: nutation turning point

At a nutation turning point,

1. 𝜃 = 0
2. ϕ = 0 necessarily
3. ω₃ = 0 necessarily
4. p_ψ = 0
5. gravity vanishes

Problem 10: stable steady precession

In the effective-potential picture, stable steady precession occurs at

1. a local maximum of U_eff
2. a local minimum of U_eff
3. any turning point
4. U_eff = 0 only
5. infinite energy

Problem 11: reaction torque

A rotor is forced to precess with angular velocity Ω and angular momentum L. The torque exerted by the rotor on its support is

1. Ω × L
2. −Ω × L
3. L × L
4. zero always
5. Ω ⋅ L

Problem 12: torque-free symmetric body

For an ideal torque-free symmetric body, the angle between its symmetry axis and the inertial angular-momentum vector is

1. constant
2. always zero
3. always increasing
4. always decreasing
5. undefined

Part II: Complete worked solutions

Solution 1

From

(7)

a perpendicular torque produces a perpendicular increment in angular momentum. To first order the magnitude is unchanged and the direction turns. Answer: (B).

Solution 2

The magnitude of

(8)

is

(9)

Answer: (B).

Solution 3

For rapid spin,

(10)

and gravity gives

(11)

Answer: (B).

Solution 4

The precession rate is inversely proportional to spin rate:

(12)

Doubling spin halves the precession rate. Answer: (B).

Solution 5

(13)

Answer: (C).

Solution 6

Nutation is oscillation or variation of the inclination angle

(14)

Answer: (B).

Solution 7

Because

(15)

is cyclic,

(16)

Answer: (A).

Solution 8

From

(17)

(18)

Answer: (A).

Solution 9

At a turning point of the effective one-dimensional inclination motion,

(19)

Answer: (A).

Solution 10

Stable steady precession requires a stable equilibrium of the inclination coordinate:

(20)

Thus it occurs at a local minimum. Answer: (B).

Solution 11

The torque on the rotor is

(21)

Newton’s third law gives the support torque:

(22)

Answer: (B).

Solution 12

For a torque-free symmetric body,

(23)

is fixed in inertial space and the symmetry axis precesses around it with constant inclination. Answer: (A).

2 GRE checklist

1. Start with τ = dL∕dt rather than memorizing a precession direction.
2. Use τ = Ω_p × L for steady vector precession.
3. Apply the fast-top formula only when spin angular momentum dominates.
4. Keep spin, precession, and nutation conceptually separate.
5. For the symmetric heavy top, exploit the conserved p_ϕ and p_ψ.
6. Use E = U_eff to identify nutation turning points.
7. Stable steady precession corresponds to a minimum of the effective potential.
8. Remember the equal-and-opposite reaction torque on a support that forces rotor precession.

References

References