GRE Physics Companion: Torque and Fixed-Axis Rotational Dynamics

GRE Physics Companion: Torque and Fixed-Axis Rotational Dynamics

The key fixed-axis equations are

(1)

(2)

and

(3)

1 High-value GRE facts

1. Torque magnitude is τ = rF sin ϕ.
2. The lever arm is the perpendicular distance to the force line of action.
3. A force through the axis has zero torque about that axis.
4. Fixed-axis dynamics uses ∑ τ_z = I_zα.
5. The torque axis and inertia axis must match.
6. A massive accelerating Pulley generally has unequal tensions.
7. No slip at a pulley rim gives a = αR.
8. Constant net torque and constant I imply constant angular acceleration.
9. rotational power is P = τω.
10. Torque integrated over time gives angular impulse.

Part I: Original GRE-style problems

Problem 1: torque angle

A force F is applied at distance R from an axis and makes angle 30^∘ with the radius vector. The torque magnitude is

1. 0
2. RF∕2
3. RF∕2
4. RF
5. 2RF

Problem 2: force through axis

A force has a line of action passing directly through the rotation axis. Its torque about that axis is

1. zero
2. F
3. RF
4. RF²
5. dependent only on mass

Problem 3: disk angular acceleration

A uniform disk of mass M and radius R experiences tangential rim force F. Its angular acceleration is

1. F∕(MR)
2. 2F∕(MR)
3. FR∕M
4. F∕(2MR)
5. 2FR∕M

Problem 4: ring versus disk

A ring and disk have equal M and R and experience the same rim torque. The ratio of angular accelerations is

(4)

equal to

1. 1∕4
2. 1∕2
3. 1
4. 2
5. 4

Problem 5: zero net torque

A rigid body has zero net axial torque and nonzero angular velocity. Its angular velocity is

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

Problem 6: braking torque

A rotor with moment of inertia I and angular speed ω₀ is stopped by constant braking torque magnitude τ_b. The stopping time is

1. τ_b∕(Iω₀)
2. I∕(τ_bω₀)
3. Iω₀∕τ_b
4. Iω₀²∕τ _b
5. τ_bω₀∕I

Problem 7: rotational power

A motor supplies torque τ to a shaft rotating at angular speed ω. The instantaneous mechanical power is

1. τ∕ω
2. ω∕τ
3. τω
4. τω²
5. τ²ω

Problem 8: angular impulse

The area under a torque-versus-time graph equals

1. angular acceleration
2. angular displacement
3. change in angular momentum
4. rotational kinetic energy
5. moment of inertia

Problem 9: massive pulley tensions

A massive pulley accelerates clockwise due to two tangential tensions T₁ and T₂. Which statement is generally correct?

1. T₁ = T₂ must always hold
2. the tensions must differ if they provide nonzero net pulley torque
3. both tensions must be zero
4. pulley inertia has no effect
5. angular acceleration must be zero

Problem 10: gravity torque on rod

A uniform rod of length L and mass M is horizontal and pivoted at one end. The magnitude of gravitational torque about the pivot is

1. 0
2. MgL∕4
3. MgL∕2
4. MgL
5. 2MgL

Problem 11: rod angular acceleration

For the rod in Problem 10, I_pivot = ML²∕3. Its initial angular acceleration magnitude is

1. g∕(2L)
2. g∕L
3. 3g∕(2L)
4. 2g∕L
5. 3g∕L

Problem 12: couple

Two equal and opposite parallel forces of magnitude F are separated by perpendicular distance d. Their net effect is

1. zero force and zero torque
2. force 2F and zero torque
3. zero force and torque magnitude Fd
4. force F and torque Fd∕2
5. torque F∕d

Part II: Complete worked solutions

Solution 1

(5)

Answer: (B).

Solution 2

The lever arm is zero because the force line of action passes through the axis:

(6)

Answer: (A).

Solution 3

For a disk,

(7)

The rim torque is

(8)

Therefore

α	=	(9)
	= .	(10)

Answer: (B).

Solution 4

For the ring,

(11)

For the disk,

(12)

At equal torque,

(13)

Thus

(14)

Answer: (D).

Solution 5

Zero net torque gives

(15)

Therefore the angular velocity remains constant. Answer: (D).

Solution 6

The angular deceleration magnitude is

(16)

Thus

(17)

Answer: (C).

Solution 7

For fixed-axis rotation,

(18)

Answer: (C).

Solution 8

Angular impulse is

(19)

Answer: (C).

Solution 9

A nonzero angular acceleration of a pulley with nonzero inertia requires

(20)

Thus the tensions generally differ. Answer: (B).

Solution 10

The rod center of mass lies at L∕2 from the pivot, and gravity is perpendicular to the horizontal rod:

(21)

Answer: (C).

Solution 11

α	=	(22)
	=	(23)
	= .	(24)

Answer: (C).

Solution 12

The equal and opposite forces cancel translationally, but their separated lines of action produce a couple:

(25)

Answer: (C).

2 GRE checklist

1. Choose the torque axis before computing any moment arms.
2. Use only the perpendicular force component or perpendicular lever arm.
3. Match the torque axis to the moment-of-inertia axis.
4. Treat τ = Iα as a fixed-axis scalar relation, not a universal Vector Identity.
5. Include translation when the center of mass accelerates.
6. Massive pulleys generally require different tensions.
7. Constant torque gives constant α only when I is constant.
8. Use torque-time area for angular impulse and torque-angle area for rotational work.

References

References