Questions: Instantaneous Center of Rotation Method

Question 1

A wheel of radius R rolls without slipping on a flat surface at angular velocity ω. What is the speed of the topmost point of the wheel?

• A. ωR — the same as the wheel's center, since all points of a rigid body share the same velocity
• B. 2ωR — the top is twice as far from the instantaneous center (the contact point) as the axle
• C. ωR/2 — the top's velocity is reduced by the no-slip constraint
• D. Zero — the no-slip condition makes the entire wheel instantaneously stationary

Answer: (B)

For a wheel rolling without slipping, the contact point has zero velocity (no-slip condition), making it the instantaneous center. The axle is at distance R from the IC, so v_axle = ωR. The topmost point is at distance 2R from the IC, so v_top = ω(2R) = 2ωR. This is why the top of a rolling wheel blurs in photographs while the bottom appears almost stationary — the IC method gives immediate insight that would require vector addition of translation and rotation components otherwise.

Question 2

An engineer locates the instantaneous center of a connecting rod and computes all point velocities using v = ω·r. She then computes accelerations using a = ω²·r directed toward the IC. What is wrong with this approach?

• A. Nothing — the IC method is equally valid for both velocities and accelerations at any instant
• B. The IC is a moving point with its own velocity; the simple centripetal formula a = ω²·r toward the IC does not account for this, giving incorrect accelerations
• C. She should use α·r instead of ω²·r — the angular acceleration, not angular velocity squared, gives the acceleration
• D. She should find a separate 'acceleration center' that always coincides with the velocity IC

Answer: (B)

The IC gives the correct velocity field for this instant, but the IC itself is generally moving — it has a non-zero velocity as the mechanism evolves. Computing acceleration requires accounting for this motion of the IC. For a purely rotating body about a truly fixed axis, a = ω²·r works. For a body in general plane motion, the acceleration of any point involves both the centripetal acceleration and Coriolis-like terms arising because the reference point (the IC) is accelerating. Using only a = ω²·r toward the IC introduces error in all but the simplest special cases.

Question 3

The instantaneous center of rotation should generally lie within the physical boundary of the moving body.

• T. True
• F. False

Answer: False

The IC is a geometric point determined by the intersection of perpendiculars to known velocity directions. It can lie anywhere in the plane — including far outside the body. For example, a rod with one end constrained to slide horizontally and the other vertically has an IC that traces a circle far outside the rod itself. For a body in near-pure-translation (very small ω), the IC approaches infinity. What matters is the mathematical construction, not whether the IC corresponds to a physical material point of the body.

Question 4

For any point on a rigid body undergoing plane motion, if the instantaneous center is known, the velocity of that point is perpendicular to the line from the IC to the point, with magnitude equal to ω times the distance from the IC.

• T. True
• F. False

Answer: True

This follows directly from the definition: since the body's motion is instantaneously equivalent to pure rotation about the IC (a point with zero velocity), every point obeys the pure-rotation relationship v = ω·r with direction perpendicular to the radius from IC. This is the computational power of the method — once the IC is located, all velocity calculations reduce to this simple relationship without decomposing motion into translational and rotational components separately.

Question 5

Why can the instantaneous center of rotation be used to find velocities but not accelerations of points on a rigid body in plane motion?

Model answer: The IC gives a correct snapshot of the velocity field at one instant by treating the motion as instantaneous pure rotation. However, the IC is not a fixed point — it moves continuously as the body moves, tracing a path called the centrode. Because the IC itself has a non-zero velocity (and acceleration), the acceleration field of the body is not simply centripetal acceleration directed toward the IC. The correct acceleration computation requires the full kinematic equations including the acceleration of the reference point, producing additional terms that the simple a = ω²·r formula misses. The IC is a snapshot tool: exact for velocities at one moment, insufficient for accelerations.

This limitation is frequently overlooked by students who, having successfully used the IC for velocities, attempt to extend it to accelerations. The centrode (locus of the IC over time) has its own acceleration, and that acceleration couples back into the body's acceleration field. For accelerations, the safest approach remains the standard kinematic equation a_B = a_A + α × r_{A/B} − ω²·r_{A/B}.