Statics and Dynamics Quiz: Instantaneous Center Of Zero Velocity
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Instantaneous Center Of Zero VelocityQuestion 1 of 7

A rigid body undergoes general planar motion. At a particular instant, two points on the body, M and N, have velocity vectors that are parallel to each other (both pointing in the same direction) but not equal in magnitude.

What can be correctly concluded about the instantaneous center of zero velocity (IC) of this rigid body at that instant?

The IC lies at infinity along the direction of the velocity vectors themselves, because parallel velocities of different magnitudes imply that the perpendiculars to those velocities are also parallel and never intersect at a finite point.
The IC does not exist at a finite location; instead, it lies at infinity in the direction perpendicular to the velocity vectors, because the body must be in pure translation when two points share the same velocity direction.
The IC exists at a finite location off the line MN, found by drawing perpendiculars to the velocity vectors at M and N, which are parallel lines meeting at a point that is offset laterally from MN.
The IC exists at a finite location and lies on the line MN, because parallel but unequal velocities at two points require the IC to be collinear with those points at a finite distance determinable by similar triangles.
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Statics and Dynamics Quiz

Statics and Dynamics Quiz: Instantaneous Center Of Zero Velocity

Practice Instantaneous Center Of Zero Velocity in Statics and Dynamics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Instantaneous Center Of Zero Velocity, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics and Dynamics.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A rigid body undergoes general planar motion. At a particular instant, two points on the body, M and N, have velocity vectors that are parallel to each other (both pointing in the same direction) but not equal in magnitude.

What can be correctly concluded about the instantaneous center of zero velocity (IC) of this rigid body at that instant?

  1. The IC lies at infinity along the direction of the velocity vectors themselves, because parallel velocities of different magnitudes imply that the perpendiculars to those velocities are also parallel and never intersect at a finite point.
  2. The IC does not exist at a finite location; instead, it lies at infinity in the direction perpendicular to the velocity vectors, because the body must be in pure translation when two points share the same velocity direction.
  3. The IC exists at a finite location off the line MN, found by drawing perpendiculars to the velocity vectors at M and N, which are parallel lines meeting at a point that is offset laterally from MN.
  4. The IC exists at a finite location and lies on the line MN, because parallel but unequal velocities at two points require the IC to be collinear with those points at a finite distance determinable by similar triangles. (correct answer)
Explanation: Whenever you encounter a question about instantaneous centers (IC), your go-to tool is this rule: the IC lies at the intersection of the perpendiculars drawn to the velocity vectors at each point. This is the geometric foundation for locating the IC in any planar motion problem. Here, points M and N have velocities that are parallel (same direction, different magnitudes). The perpendiculars to parallel vectors are themselves parallel — but "parallel" in this context means they both point in the same perpendicular direction, not that they run along the same line. Since M and N are distinct points on the body, those perpendiculars are offset from each other, running parallel along the line MN itself. Two parallel lines that are collinear (lying on the same line) don't fail to intersect — they overlap entirely, meaning every point on that line is a candidate. The IC is then pinned to a specific finite point on line MN by applying similar triangles: since velocity magnitude is proportional to distance from the IC, and the magnitudes differ, the IC sits at a calculable finite location along MN. This confirms D is correct. A is wrong because perpendiculars to the velocities run along MN, not in the direction of the velocities themselves — the geometry is misidentified. B is wrong on two counts: pure translation requires equal velocities (same magnitude and direction), and different magnitudes rule that out; additionally, the IC doesn't vanish to infinity here. C is wrong because the perpendiculars aren't offset laterally — they're collinear along MN, not parallel lines separated in space. Your study tip: always sketch the perpendiculars. When velocities are parallel, ask yourself whether the perpendiculars are truly separate parallel lines (IC at infinity) or collinear along the connecting line (IC finite, found by similar triangles).

Question 2

A rigid disk of radius RR rolls without slipping inside a fixed circular track of radius 2R2R. The center of the disk moves along a circle of radius RR (since the inner disk radius is RR and the outer fixed circle radius is 2R2R, the center traces a circle of radius RR). At a particular instant, the center C of the rolling disk is at the rightmost point of its circular path (i.e., C is directly to the right of the center of the fixed track, O).

At this instant, where is the instantaneous center of zero velocity of the rolling disk?

  1. At the center O of the fixed circular track, because the contact point between the rolling disk and the fixed track is the IC, which is at distance 2R2R from O, but the geometry of internal rolling places the effective IC at O for this specific radius ratio of 2:1.
  2. At the contact point between the rolling disk and the fixed circular track, located on the right side of the fixed track at distance 2R2R from O, because rolling without slipping always places the IC at the contact point. (correct answer)
  3. At the center C of the rolling disk, because for internal rolling, the curvature of the contact path causes the IC to coincide with the moving disk's center rather than the contact point.
  4. At the leftmost point of the rolling disk (the point on the disk diametrically opposite the contact point), because internal rolling reverses the IC location compared to external rolling on a flat surface.
Explanation: Whenever you see a problem involving rolling without slipping, your first instinct should be to locate the instantaneous center of zero velocity (IC). The IC is the point on (or associated with) a rigid body that has zero velocity at that instant — and for any body rolling without slipping on a surface, that point is always the contact point, because no slipping means the contact point shares the velocity of the surface it rolls against (zero, since the track is fixed). This principle holds regardless of whether the rolling is on a flat surface, on the outside of a curve, or on the inside of a curve. When the disk rolls inside the fixed circular track of radius 2R2R, the contact point lies on the right side of the fixed track, at a distance 2R2R from O. Since the track is stationary, that contact point has zero velocity — making it the IC. Answer B is correct. Answer A is wrong because it invents a special exception for the 2:1 radius ratio. No such exception exists; the IC is not relocated to O by geometric coincidence. Answer C is wrong because the disk's center C is clearly moving along its circular path and therefore has nonzero velocity — it cannot be the IC. Answer D is wrong because internal rolling does not "reverse" the IC location. The diametrically opposite point to the contact point is actually the point of maximum speed on the disk, not zero speed. As a study tip: never let the word "internal" trick you into abandoning the fundamental rule. Rolling without slipping always places the IC at the contact point, full stop.

Question 3

A ladder of length LL leans against a smooth vertical wall, with its base on a smooth horizontal floor. Both surfaces are frictionless. The base is given an initial push, and the ladder subsequently slides freely. At a certain instant, the base A has velocity vAv_A directed horizontally away from the wall, and the tip B has velocity vBv_B directed vertically downward. The ladder makes angle θ\theta with the horizontal.

A student claims: 'Because both surfaces are frictionless, the ladder falls faster than it would if there were friction, so the instantaneous center of the ladder moves closer to the ladder's midpoint compared to the frictionful case.' Evaluate this claim using IC analysis.

  1. The claim is correct. With frictionless surfaces, the normal forces at A and B are both perpendicular to the surfaces, causing a larger angular acceleration, which shifts the IC toward the midpoint of the ladder compared to the frictional case.
  2. The claim is incorrect. The IC location depends only on the instantaneous velocity directions of A and B, not on whether friction is present. The IC is always at the intersection of the vertical through A and the horizontal through B, regardless of the friction condition, and its position relative to the midpoint depends only on θ\theta. (correct answer)
  3. The claim is incorrect. Frictionless surfaces prevent any tangential force at the contacts, which means the IC is always located at the center of mass of the ladder, regardless of the angle θ\theta.
  4. The claim is partially correct. The IC location is independent of friction for the instantaneous velocity field, but friction affects the acceleration field, which in turn shifts the IC to a different position in the next instant, effectively moving it closer to the midpoint over time.
Explanation: When analyzing rigid body kinematics, you must carefully distinguish between the instantaneous velocity field and the force/acceleration analysis — they follow different rules. The Instantaneous Center (IC) of a rigid body is defined purely kinematically: it is the point with zero velocity at that instant. For the sliding ladder, point A can only move horizontally (wall constrains vertical motion — frictionless or not, the wall still exerts a normal force), and point B can only move vertically (floor's normal force keeps it in contact). Therefore, the velocity of A is always horizontal and the velocity of B is always vertical. The IC must lie on the vertical line through A and the horizontal line through B — their intersection. This geometric construction depends entirely on the contact geometry and the resulting velocity directions, not on whether friction exists. The IC position relative to the ladder depends only on θ\theta, giving coordinates that follow directly from trigonometry. Answer B is correct for exactly this reason: IC location is a kinematic quantity determined by instantaneous velocity directions, and those directions are fixed by the constraint surfaces regardless of surface friction. Answer A is wrong because it conflates angular acceleration (a force-dependent quantity) with IC location (a velocity-dependent quantity). Larger angular acceleration does not "shift" the IC. Answer C is wrong because the IC being at the center of mass is a special condition, not a general consequence of frictionless surfaces. This confuses force conditions with velocity geometry. Answer D is wrong because while friction does affect how velocities evolve (changing θ\theta and hence IC position over time), the IC at any given instant is still determined solely by instantaneous velocities — not by friction's influence on the next instant. Study tip: On dynamics exams, always ask yourself: "Am I dealing with a velocity question or a force/acceleration question?" IC analysis belongs strictly to the velocity domain.

Question 4

A rigid bar PQ of length 2 m2\text{ m} moves in the plane. At a certain instant, point P has velocity v⃗P=4i^+3j^ m/s\vec{v}_P = 4\hat{i} + 3\hat{j}\text{ m/s} and point Q has velocity v⃗Q=4i^−3j^ m/s\vec{v}_Q = 4\hat{i} - 3\hat{j}\text{ m/s}.

What can be determined about the instantaneous center of zero velocity of bar PQ at this instant?

  1. The IC lies on the perpendicular bisector of PQ at a finite distance from the midpoint, because the equal x-components of velocity indicate that the IC must be equidistant from P and Q in the horizontal direction.
  2. The IC lies at infinity in the i^\hat{i} direction, because the equal x-components of v⃗P\vec{v}_P and v⃗Q\vec{v}_Q indicate that PQ undergoes pure translation in the x-direction at this instant.
  3. The IC lies on the line PQ itself at a finite location, found by drawing perpendiculars to v⃗P\vec{v}_P and v⃗Q\vec{v}_Q at P and Q respectively and finding their intersection, which falls on PQ because the relative velocity of P with respect to Q is perpendicular to PQ. (correct answer)
  4. The IC cannot be determined from this information alone because the length of the bar (2 m2\text{ m}) is needed along with the velocity components to locate the IC using the angular velocity formula ω=Δv/L\omega = \Delta v / L.
Explanation: When analyzing rigid body motion, the instantaneous center of zero velocity (IC) is found at the intersection of lines drawn perpendicular to the velocity vectors at each point. This geometric method works regardless of whether you know the bar's orientation explicitly — the velocity directions tell you everything. Here, v⃗P=4i^+3j^\vec{v}_P = 4\hat{i} + 3\hat{j} and v⃗Q=4i^−3j^\vec{v}_Q = 4\hat{i} - 3\hat{j}. Notice the velocities are mirror images across the x-axis: equal horizontal components, opposite vertical components. The perpendicular to v⃗P\vec{v}_P points in the direction 3i^−4j^3\hat{i} - 4\hat{j}, and the perpendicular to v⃗Q\vec{v}_Q points in the direction 3i^+4j^3\hat{i} + 4\hat{j}. These two perpendicular lines intersect at a finite point along the bar itself, confirming C is correct. Physically, the relative velocity v⃗P/Q=6j^\vec{v}_{P/Q} = 6\hat{j} is purely vertical, meaning it's perpendicular to any horizontal line — so PQ must be oriented horizontally, and the IC lies on that horizontal line between (or beyond) P and Q. Choice A confuses "equidistant in the horizontal direction" with the actual perpendicular-bisector construction — the IC location depends on velocity directions, not just component magnitudes. Choice B is the most tempting trap: equal x-components do not mean pure translation. Pure translation requires identical full velocity vectors; here the y-components differ, so rotation is occurring. Choice D is simply false — the IC location requires velocity directions (already given), not the bar length. Length LL helps find ω\omega, but that's a separate calculation. Study tip: Always check all components before concluding pure translation. If any component differs between two points, the body is rotating — find the IC using perpendiculars to the full velocity vectors.

Question 5

A uniform disk of radius RR rolls without slipping on a flat horizontal surface. A point P is located on the rim of the disk. At the instant when P is at the topmost position (directly above the disk's center), the center C of the disk moves with velocity vCv_C to the right.

Which of the following correctly describes the velocity of point P and the location of the instantaneous center of zero velocity (IC) at this instant?

  1. Point P has speed vCv_C directed to the right, and the IC is located at the contact point between the disk and the ground, since P is diametrically opposite the IC and the velocity magnitude equals ωR\omega R.
  2. Point P has speed 2 vC\sqrt{2}\,v_C directed at 45° above horizontal, and the IC is located at the center C of the disk, since C is the reference point for computing all velocities on the disk.
  3. Point P has speed 2vC2v_C directed upward, and the IC is located at the contact point between the disk and the ground, since the instantaneous center is always directly below the center for a rolling disk.
  4. Point P has speed 2vC2v_C directed to the right, and the IC is located at the contact point between the disk and the ground, since rolling without slipping requires zero velocity at the contact point. (correct answer)
Explanation: When a disk rolls without slipping, the key concept to anchor everything is the instantaneous center of zero velocity (IC). For any rolling disk, the no-slip condition guarantees that the contact point has zero velocity at that instant — making it the IC. Every other point on the disk rotates about this IC with angular velocity ω\omega. Since the center C moves at vCv_C and sits a distance RR above the IC, you can find ω\omega: vC=ωRv_C = \omega R, so ω=vC/R\omega = v_C / R. Point P sits at the top of the disk, directly above C, meaning it is a distance 2R2R from the IC (the contact point). Its speed is therefore vP=ω⋅2R=vCR⋅2R=2vCv_P = \omega \cdot 2R = \frac{v_C}{R} \cdot 2R = 2v_C. The direction of P's velocity is always perpendicular to the line connecting P to the IC — since P is directly above the IC, that line is vertical, so P moves horizontally to the right. This confirms D. A gets the IC location right but claims P's speed equals vCv_C, ignoring that P is twice as far from the IC as the center is. B places the IC at the center C, which is wrong — C itself has nonzero velocity, so it cannot be the IC. It also produces an incorrect speed and direction. C correctly states P's speed as 2vC2v_C and correctly identifies the IC, but claims the velocity is directed upward — this is the critical error, since the velocity must be perpendicular to the vertical line from IC to P, meaning it points horizontally. Your study tip: always determine the IC first, then compute velocity as v=ω⋅dv = \omega \cdot d, directed perpendicular to the line from the IC to the point. The direction trap in C is very common on dynamics exams.

Question 6

A four-bar linkage consists of fixed ground link OO', crank OA of length rr, coupler AB of length ℓ\ell, and rocker O'B of length ss. At a particular instant, the crank OA is perpendicular to the ground link OO' (i.e., OA points straight up), and the coupler AB is horizontal. The angular velocity of the crank is ωOA\omega_{OA}.

To find the angular velocity of the rocker O'B using the instantaneous center of zero velocity of coupler AB, a student correctly identifies the IC of AB as point I. Which of the following correctly describes the subsequent steps and result?

  1. The student computes vA=ωOA⋅rv_A = \omega_{OA} \cdot r directed horizontally, locates I at the intersection of the vertical through A and the line through B along O'B, then calculates ωAB=vA/(IA)\omega_{AB} = v_A / (IA) and finally ωO′B=vB/(O′B)=ωAB⋅(IB)/(O′B)\omega_{O'B} = v_B / (O'B) = \omega_{AB} \cdot (IB) / (O'B), where IA and IB are distances from I to A and B respectively. (correct answer)
  2. The student computes vA=ωOA⋅rv_A = \omega_{OA} \cdot r directed horizontally, then applies the formula ωO′B=vA⋅r/(ℓ⋅s)\omega_{O'B} = v_A \cdot r / (\ell \cdot s), since in a four-bar linkage the output angular velocity is always inversely proportional to the product of the coupler length and rocker length.
  3. The student locates I on the coupler AB between A and B, then determines ωO′B\omega_{O'B} by applying the distance ratio directly to the crank speed: ωO′B=ωOA⋅(IA/IB)\omega_{O'B} = \omega_{OA} \cdot (IA/IB), bypassing the need to compute vBv_B or account for the rocker length ss.
  4. The student notes that since OA is perpendicular to the ground and AB is horizontal, the perpendiculars to both vAv_A and vBv_B are both vertical, placing the IC of AB at infinity; this means AB translates horizontally, so vB=vAv_B = v_A and ωO′B=vA/s\omega_{O'B} = v_A / s.
Explanation: When solving for angular velocities in a four-bar linkage using the instantaneous center (IC) method, your goal is to find the point on the coupler AB that has zero velocity at that instant, then use it as a pivot to relate velocities at A and B geometrically. The IC of a link is found at the intersection of lines drawn perpendicular to the velocity at each end. At point A, the velocity vA=ωOA⋅rv_A = \omega_{OA} \cdot r is directed horizontally (since OA is vertical), so the perpendicular to vAv_A is vertical through A. At point B, since O'B is a rocker pinned at O', point B moves perpendicular to O'B — so the perpendicular to vBv_B lies along the line O'B. The IC is where these two lines intersect. From there, ωAB=vA/IA\omega_{AB} = v_A / IA, giving vB=ωAB⋅IBv_B = \omega_{AB} \cdot IB, and finally ωO′B=vB/s\omega_{O'B} = v_B / s. This is precisely what A describes — making it correct. B is wrong because no universal formula ωO′B=vA⋅r/(ℓ⋅s)\omega_{O'B} = v_A \cdot r / (\ell \cdot s) governs all four-bar linkages; output velocity depends on geometry at each instant, not a fixed product rule. C is wrong on two counts: the IC is generally not between A and B, and you cannot bypass the rocker length ss — ωO′B\omega_{O'B} requires dividing vBv_B by ss, not a simple crank-speed ratio. D is tempting but wrong. The perpendicular to vBv_B runs along O'B, which is generally not vertical — so the two perpendiculars are not parallel, and the IC is not at infinity. As a strategy, always draw the perpendiculars to velocity at both ends of the coupler separately before concluding anything about the IC's location — don't assume symmetry from the linkage's appearance.

Question 7

A rigid rod AB of length L=2 mL = 2\text{ m} moves in the plane. End A is constrained to slide along a frictionless horizontal surface, and end B is constrained to slide along a frictionless vertical wall. At the instant shown, end A has velocity vA=3 m/sv_A = 3\text{ m/s} directed horizontally (away from the wall), and the rod makes an angle of θ=30°\theta = 30° with the horizontal.

Using the instantaneous center of zero velocity (IC), what is the angular velocity ω\omega of the rod at this instant, and in which direction does it rotate?

  1. ω=3.0 rad/s\omega = 3.0\text{ rad/s}, rotating clockwise, found by locating the IC at the intersection of perpendiculars to vAv_A and vBv_B, which gives a perpendicular distance from IC to A equal to Lsin⁡θ=1.0 mL\sin\theta = 1.0\text{ m}. (correct answer)
  2. ω=1.5 rad/s\omega = 1.5\text{ rad/s}, rotating clockwise, found by locating the IC at the intersection of perpendiculars to vAv_A and vBv_B, which gives a perpendicular distance from IC to A equal to Lcos⁡θ=3 mL\cos\theta = \sqrt{3}\text{ m}.
  3. ω=1.5 rad/s\omega = 1.5\text{ rad/s}, rotating counterclockwise, found by noting that the IC lies above the rod, giving a perpendicular distance from IC to A equal to Lcos⁡θ=3 mL\cos\theta = \sqrt{3}\text{ m}.
  4. ω=2.6 rad/s\omega = 2.6\text{ rad/s}, rotating clockwise, found by using the full length L=2 mL = 2\text{ m} as the perpendicular distance from the IC to point A, since both endpoints are constrained.
Explanation: When a rigid body has two points with known velocity directions, your first move should be to locate the instantaneous center of zero velocity (IC) — the point about which the entire body appears to rotate at that instant. The IC lies at the intersection of lines drawn perpendicular to each velocity vector. Here, end A moves horizontally, so the perpendicular to vAv_A is vertical. End B moves vertically (constrained to the wall), so the perpendicular to vBv_B is horizontal. These two perpendiculars intersect directly above A and directly to the right of B — that intersection point is the IC, which geometrically sits at the corner formed by the wall and the floor's extension. The perpendicular distance from the IC to point A equals Lsin⁡θ=2sin⁡30°=1.0 mL\sin\theta = 2\sin30° = 1.0\text{ m}. Since vA=ω⋅rA/ICv_A = \omega \cdot r_{A/IC}, you get ω=vA/(Lsin⁡θ)=3/1.0=3.0 rad/s\omega = v_A / (L\sin\theta) = 3/1.0 = 3.0\text{ rad/s}. Because A moves away from the wall while the IC sits above-left of A, the rod rotates clockwise. That confirms answer A. Answer B uses Lcos⁡θ=3 mL\cos\theta = \sqrt{3}\text{ m} as the IC-to-A distance — this confuses the perpendicular distance to A with the horizontal projection of the rod, which would only apply to a different geometry. Answer C compounds that same geometric error and then incorrectly reverses the rotation direction. Answer D simply uses the full rod length L=2 mL = 2\text{ m} as the moment arm, ignoring that the IC is generally not located at one of the endpoints. As a rule of thumb: always sketch the perpendiculars to both velocity vectors — their intersection defines the IC, and the moment arm to each point is the straight-line distance from that point to the IC, not any component of the rod's length.