Questions: Rigid Body Planar Motion: Translation and Rotation

Question 1

A solid disk and a hoop of equal mass M and equal radius R are released from rest at the top of an inclined plane. Which reaches the bottom first?

• A. The hoop, because its mass concentrated at the rim gives it greater rotational momentum
• B. The disk, because its smaller moment of inertia means less energy goes into rotation and more into translational speed
• C. They arrive at the same time, because they have identical mass and radius
• D. The disk, because static friction acts more strongly on the hoop

Answer: (B)

Both start with the same potential energy Mgh, which must be split between translational KE (½Mv_cm²) and rotational KE (½Iω²). The disk (I = MR²/2) has a smaller moment of inertia than the hoop (I = MR²), so less energy goes into rotation and more into translation — the disk arrives faster. Equal mass and radius do not imply equal speed; the distribution of mass (captured by I) is what determines how energy is partitioned.

Question 2

A rigid body rolls without slipping down an incline from height h. How does its final translational speed compare to a point mass sliding frictionlessly down the same incline?

• A. Equal to √(2gh) — rolling objects reach the same speed as sliding point masses
• B. Less than √(2gh) — energy is split between translation and rotation, leaving less for translational speed
• C. Greater than √(2gh) — rotation adds kinetic energy to the system
• D. The comparison depends on the object's shape but not its mass

Answer: (B)

A frictionlessly sliding point mass converts all potential energy to translational KE: ½Mv² = Mgh, giving v = √(2gh). A rolling rigid body must also supply rotational KE (½Iω²), so less energy is available for translation. With v_cm = Rω, both kinetic energy terms depend on v_cm, and conservation of energy gives v_cm = √(2gh / (1 + I/MR²)) < √(2gh). The shape (via I/MR²) determines exactly how much slower it is.

Question 3

For a rigid body in planar motion, the translational equation ΣF = Ma_cm and the rotational equation Στ_cm = I_cm α are fully independent — a single applied force can contribute to at most one of these equations.

• T. True
• F. False

Answer: False

This is a key misconception. The same force can simultaneously contribute to both equations. Friction on a rolling body, for example, appears in ΣF = Ma_cm as a translational force and in Στ_cm = I_cm α as a torque (since it acts at the contact point, at distance R from the center of mass). The equations are independent in structure but not in the forces they share — this coupling is what makes rigid body problems solvable.

Question 4

The rolling-without-slipping constraint v_cm = Rω links translational and rotational motion, reducing the number of independent variables needed to describe the motion.

• T. True
• F. False

Answer: True

Without rolling-without-slipping, a rigid body has two independent degrees of freedom: translational (described by v_cm) and rotational (described by ω). The geometric constraint v_cm = Rω ties these together, reducing to one independent variable. This means you need one fewer equation to solve the problem. When slipping occurs, the constraint breaks, and you must treat both equations fully independently, using kinetic friction as the link between the translational and rotational dynamics.

Question 5

Explain the decomposition theorem for planar rigid body motion and why it means you need two separate equations (not one) to fully describe the motion.

Model answer: Any planar rigid body motion can be decomposed exactly into (1) translation of the center of mass, governed by ΣF = Ma_cm, and (2) rotation about the center of mass, governed by Στ_cm = I_cm α. These are two separate physical processes that happen simultaneously — the same forces appear in both equations but describe different aspects of the motion. One equation cannot capture both: Newton's second law for a point mass describes translation, but says nothing about how the body spins.

The decomposition is what makes rigid body analysis tractable. Rather than tracking every particle in the body, you track two quantities: where the center of mass goes (translation) and how the body rotates about it. The kinetic energy formula KE = ½Mv_cm² + ½I_cm ω² reflects this same decomposition — two additive terms, one for each mode of motion.