Questions: Rigid Body Kinetics — Force and Acceleration

Question 1

A uniform disk rolls without slipping down an inclined plane. A student writes the friction force as f = µ_k × N in the equations of motion. What error has the student made?

• A. The student should use µ_s instead of µ_k, since the disk is not slipping
• B. Friction at a no-slip rolling contact is an unknown to be solved from the equations of motion, not a known force equal to µN
• C. The normal force N must be resolved differently on an inclined surface before multiplying by µ
• D. Static friction is zero when a disk rolls without slipping, so the term should be omitted entirely

Answer: (B)

For rolling without slip, friction is a static friction force — it takes whatever value is required to enforce the kinematic constraint (zero slip at the contact point). This value is not µ_s×N or µ_k×N; those are the *limits* of static and kinetic friction, respectively. The actual friction force is an unknown that emerges from solving the three equations of motion simultaneously with the rolling constraint a_G = α·r. After solving, you check whether the required friction is less than µ_s×N to verify the no-slip assumption is valid. Assuming f = µN at the outset is incorrect and gives the wrong answer.

Question 2

A uniform rod is pinned at one end (fixed axis rotation) and swings from rest. As it swings, it has angular velocity ω and angular acceleration α. A student accounts for the tangential acceleration of the mass center (r·α) when writing force equations, but gets incorrect pin reactions. What term was likely forgotten?

• A. The weight of the rod, which acts at the mass center and must appear in the free-body diagram
• B. The normal (centripetal) acceleration of the mass center (ω²·r directed toward the pin), which contributes to pin reaction forces
• C. The moment of inertia, which should be computed about the pin rather than the mass center
• D. The applied moment equation, which must always be written before the force equations for rotating bodies

Answer: (B)

For fixed-axis rotation, the mass center moves in a circle. It therefore has two acceleration components: tangential (r·α, perpendicular to the radius) from angular acceleration, and normal (ω²·r, directed toward the pin) from the existing angular velocity. The normal component is often forgotten because it is zero at the instant of release from rest (ω = 0), but it grows as the rod gains angular velocity. Neglecting ω²·r means the pin reactions are calculated incorrectly at any point after motion begins.

Question 3

When taking moments about the mass center G of a rigid body in planar motion, the rotational equation of motion is ΣM_G = I_G·α, with no additional correction terms needed.

• T. True
• F. False

Answer: True

Summing moments about the mass center G is the 'clean' choice for the rotational equation because it eliminates the transport term. When you sum moments about any other point P, you must include the moment of the m·a_G vector about P as an additional term: ΣM_P = I_G·α + (r_{G/P} × m·a_G). This transport term is zero when P = G (the moment arm is zero). Many errors in rigid body kinetics come from summing moments about a convenient point like the contact patch and forgetting to add this term.

Question 4

For a rigid body in pure translation (α = 0), the net moment about the mass center is expected to be zero, which means no individual forces can create moments about G.

• T. True
• F. False

Answer: False

Pure translation means α = 0, so the moment equation gives ΣM_G = I_G·(0) = 0 — the net moment about G is zero. But individual forces can still create nonzero moments that happen to cancel. For example, in a block sliding on a surface, gravity acts downward at G, the normal force acts upward at the contact surface below G, and friction acts horizontally at the contact. Each creates a moment about G, but they sum to zero for pure translation. The moment equation in this case constrains where the resultant force must act, not that no forces have moment arms.

Question 5

When analyzing a rigid body rolling without slip, why must friction be treated as an unknown to solve for rather than set equal to µN? Walk through the correct procedure.

Model answer: Friction at a no-slip rolling contact is static friction, which adjusts to whatever value is needed to prevent slipping — it can be anywhere from zero to µ_s×N. Its actual value depends on the applied forces, the geometry, and the inertia of the body. The correct procedure: (1) Write the three equations of motion (ΣF_x = m·a_G, ΣF_y = m·a_G, ΣM_G = I_G·α) with friction f as an unknown. (2) Apply the rolling constraint: a_G = α·r, giving a fourth equation. (3) Solve the system for all unknowns including f. (4) Check: if |f| ≤ µ_s×N, rolling without slip is valid. If |f| > µ_s×N, the body actually slips — restart with f = µ_k×N as a known and the rolling constraint removed.

The procedure reveals why the no-slip check comes at the end: you need to know what friction the no-slip condition demands before you can check whether it is physically achievable. Assuming f = µN at the start is incorrect because it imposes a specific friction value that may or may not match what the rolling constraint requires.