GRE Physics Companion: Fixed-Axis Rotation and Particle Kinematics
For a material point at perpendicular radius ρ,
and
The corresponding magnitudes are
Figure 1. A compact strategy for fixed-axis particle kinematics: identify perpendicular radius,
solve the common angular motion, and then compute the point’s tangential and Normal motion.
1 High-value GRE facts
- Use perpendicular distance to the axis, not distance to an arbitrary origin.
- All points share the same ω and α, but not the same v or a.
- velocity is tangent to the circular path.
- normal acceleration points toward the axis.
- tangential acceleration changes the speed.
- an contains ω2.
- A point displaced along the axis has the same kinematics as a point with the same
perpendicular radius.
- The rotating basis vectors must be differentiated.
- The double-cross-product acceleration is inward.
- Relative velocity between two points is perpendicular to their separation vector.
Part I: Original GRE-style problems
Problem 1: speed at a known radius
A point lies 0.30 m from a fixed axis and the body rotates at 10 rad∕s. The point’s speed
is
- 0.3 m∕s
- 1.0 m∕s
- 3.0 m∕s
- 10 m∕s
- 30 m∕s
Problem 2: normal acceleration
For the point in Problem 1, the normal acceleration magnitude is
- 3 m∕s2
- 10 m∕s2
- 30 m∕s2
- 100 m∕s2
- 300 m∕s2
Problem 3: tangential acceleration
If the same body has α = 4 rad∕s2, the tangential acceleration magnitude at ρ = 0.30 m
is
- 0.3 m∕s2
- 0.75 m∕s2
- 1.2 m∕s2
- 4.0 m∕s2
- 12 m∕s2
Problem 4: two radii
Two points on the same rigid body lie at radii r and 4r. Their speed ratio is
- 1∕4
- 1
- 2
- 4
- 16
Problem 5: same two radii
For the points in Problem 4, their normal-acceleration ratio at the same instant is
- 1∕4
- 1
- 2
- 4
- 16
Problem 6: axial offset
Two material points have equal perpendicular radius ρ but different positions along the fixed axis.
Their speed magnitudes are
- always equal
- proportional to their axial separation
- inversely proportional to their axial separation
- equal only if α = 0
- zero
Problem 7: direction of velocity
The instantaneous velocity of a point in fixed-axis rotation is
- radial outward
- radial inward
- tangent to the circular path
- parallel to the axis
- always zero
Problem 8: constant angular speed
If ω is constant and nonzero, the particle acceleration is
- zero
- purely tangential
- purely inward normal
- parallel to velocity
- outward
Problem 9: vector velocity
If ω = 5ez rad∕s and r = 2ex m, then
- v = 10ex m∕s
- v = −10ex m∕s
- v = 10ey m∕s
- v = −10ey m∕s
- v = 0
Problem 10: relative velocity rigidity check
For two points A and B on the same rigid body,
is
- positive
- negative
- zero
- equal to ω
- equal to α
Problem 11: inverse relation
A point at radius 0.25 m has tangential speed 5 m∕s. The angular speed magnitude
is
- 1.25 rad∕s
- 5 rad∕s
- 10 rad∕s
- 20 rad∕s
- 25 rad∕s
Problem 12: kinetic-energy bridge
For one particle in fixed-axis rotation, its kinetic energy can be written
- mρω
mρω2
mρ2ω2
- mρ2ω
mω2∕ρ
Part II: Complete worked solutions
Solution 1
Answer: (C).
Solution 2
Answer: (C).
Solution 3
Answer: (C).
Solution 4
Since v = ρ|ω| and both points share the same ω,
Answer: (D).
Solution 5
Since an = ρω2 at the same instant,
Answer: (D).
Solution 6
Fixed-axis kinematics depends on perpendicular radius, not axial position. Equal ρ and common ω
give equal speeds. Answer: (A).
Solution 7
The velocity is in the e𝜃 direction, tangent to the circular path. Answer: (C).
Solution 8
Constant angular speed implies α = 0, so only
remains. Answer: (C).
Solution 9
| v | = ω × r | (11)
|
| = 5ez × 2ex | (12)
|
| = 10ey m∕s. | (13) |
Answer: (C).
Solution 10
Rigid body relative velocity is perpendicular to separation:
Answer: (C).
Solution 11
Answer: (D).
Solution 12
Since v = ρ|ω|,
Answer: (C).
2 GRE checklist
- Identify the perpendicular radius first.
- Keep er and e𝜃 directions straight.
- Velocity is tangential, not radial.
- Use an = ρω2 and a
t = ρ|α|.
- Equal angular speed does not imply equal linear speed at different radii.
- Axial offset does not change fixed-axis kinematics at the same ρ.
- Use cross products for vector direction questions.
- Check rigidity with rB∕A ⋅ vB∕A = 0.
References
References
[1] J. R. Taylor, Classical Mechanics, University Science Books, 2005.
[2] D. Kleppner and R. Kolenkow, An Introduction to Mechanics, 2nd ed., Cambridge
University Press, 2014.
[3] OpenStax, University Physics, Volume 1, Rice University, 2016.