Instantaneous centres and relative acceleration

Lesson 8 found velocities in a linkage by writing \(\mathbf{v}_B = \mathbf{v}_A + \boldsymbol{\omega}\times\mathbf{r}_{B/A}\) and solving the components. This lesson adds two tools. The instantaneous centre of zero velocity is a geometric shortcut that often finds velocities in a few lines, and it builds a feel for how a mechanism moves that equations alone do not give. The relative acceleration equation extends the analysis to accelerations, which is what you need for forces: the second law cares about accelerations, not velocities. The lesson finishes with a short look at motion relative to rotating axes, which explains where the Coriolis acceleration of Lesson 2 comes from in rigid-body terms.

The instantaneous centre of zero velocity

At any instant, a rigid body in general plane motion has velocities that look exactly as if the body were rotating about some point, the instantaneous centre of zero velocity, often written IC. That point may lie on the body or off it, in empty space beside the mechanism. At that instant its velocity is zero, and every point on the body moves at right angles to the line joining it to the IC, with a speed equal to \(\omega\) times its distance from the IC:

The IC exists because the relative velocity equation can be applied with any reference point. If we pick as the reference the one point whose velocity happens to be zero, the translation part vanishes and only rotation is left. As the body moves, the IC moves too, which is why it is called instantaneous: it is the centre of rotation only for that instant.

Locating the IC uses the fact that each point's velocity is perpendicular to the line from the IC. Three cases cover almost every problem.

• If the directions of the velocities of two points are known, draw lines through each point perpendicular to its velocity. The IC is where the lines cross.
• If the velocities of two points are parallel and perpendicular to the line joining them, the IC lies on that line, located by proportion from the two speeds. If the speeds are equal, the IC is infinitely far away and the body is translating at that instant, with \(\omega = 0\).
• A wheel that rolls without slipping has its IC at the contact point, because the contact point and the ground have the same velocity, zero.

The IC method also explains the special positions of the mechanism at a glance. At 90°, the crank is vertical, the perpendicular to \(\mathbf{v}_B\) is horizontal and never meets the vertical through \(C\): the IC is at infinity, the rod is translating, and the slider moves at exactly the crank pin speed, 2.0 m/s. At 0° and 180°, the crank and rod are in line, the IC of the rod coincides with \(C\), and the slider is momentarily at rest. These are the dead-centre positions.

Rolling without slipping

A wheel of radius \(r\) rolling without slipping on a flat surface has its IC at the contact point. Its centre moves at \(v_G = \omega r\), and since that holds at every instant, differentiating gives \(a_G = \alpha r\) for the centre's acceleration along the surface. The top of the wheel, twice as far from the IC as the centre, moves at \(2v_G\). This is why the top of a bicycle tyre looks blurred in photographs while the bottom looks sharp.

That warning is the single most important limitation of the instantaneous centre. It gives velocities correctly at an instant, but because the IC itself moves from instant to instant, treating it as a fixed pivot for accelerations gives wrong answers. For accelerations, go back to the relative acceleration equation.

Relative acceleration

Differentiating the relative velocity equation for two points on the same rigid body gives

The relative acceleration of \(B\) about \(A\) has two parts, exactly like a point on a body rotating about a fixed axis through \(A\): a tangential part \(\alpha\,AB\), perpendicular to \(AB\), and a normal part \(\omega^2 AB\), directed from \(B\) toward \(A\). The angular velocity \(\omega\) is already known from the velocity analysis, so the normal part is fully known. The unknowns are usually the angular acceleration \(\alpha\) of a link and the magnitude of the acceleration of a point whose path is known, such as a slider. The vector equation gives two scalar equations, enough for both.

A practical order of work: always complete the velocity analysis first, because the accelerations need the angular velocities. Then write the acceleration equation link by link, starting from a point whose acceleration is known (a fixed pivot, or a crank pin on a crank turning at a known rate).

The program below repeats the velocity and acceleration analysis round half a revolution and checks the slider acceleration against a numerical second derivative of the position formula from Lesson 8. It also confirms, by an internal check, that the rod's angular velocity agrees with the instantaneous centre construction at every angle where the IC is at a finite distance.

# Pusher slider-crank (r = 0.10 m, L = 0.30 m, crank at a constant
# 20 rad/s CCW). Velocities from the instantaneous centre of the rod,
# accelerations from the relative-acceleration equation, checked against
# a numerical second derivative of the slider position.
import math

r, L, w = 0.10, 0.30, 20.0

def xC(th):
    return r * math.cos(th) + math.sqrt(L**2 - (r * math.sin(th))**2)

def analyse(th):
    B = (r * math.cos(th), r * math.sin(th))
    C = (xC(th), 0.0)
    rCB = (C[0] - B[0], C[1] - B[1])
    vB = (-w * B[1], w * B[0])
    # Rod angular velocity from the j-component of vC = vB + wBC k x rCB
    wBC = -vB[1] / rCB[0]
    # Check with the instantaneous centre: crank line meets vertical through C
    if abs(math.cos(th)) > 1e-9:
        I = (C[0], C[0] * math.tan(th))
        IB = math.hypot(B[0] - I[0], B[1] - I[1])
        assert abs(abs(wBC) - r * w / IB) < 1e-9
    vC = vB[0] - wBC * rCB[1]
    # Accelerations: aB = -w^2 rB (alpha_crank = 0)
    aB = (-w * w * B[0], -w * w * B[1])
    # aC = aB + aBC k x rCB - wBC^2 rCB, with aC purely along x
    aBC = -(aB[1] - wBC**2 * rCB[1]) / rCB[0]
    aC = aB[0] - aBC * rCB[1] - wBC**2 * rCB[0]
    return wBC, vC, aBC, aC

h = 1e-4
print(" theta  w_BC(rad/s)  v_C(m/s)  alpha_BC(rad/s^2)  a_C(m/s^2)  a_C numerical")
for deg in (0, 30, 60, 90, 120, 150, 180):
    th = math.radians(deg)
    wBC, vC, aBC, aC = analyse(th)
    num = (xC(th + h) - 2 * xC(th) + xC(th - h)) / h**2 * w * w
    print(f"{deg:5d}  {round(wBC, 3) + 0.0:10.3f}  {round(vC, 3) + 0.0:8.3f}  {round(aBC, 1) + 0.0:15.1f}  {aC:10.2f}  {num:10.2f}")

 theta  w_BC(rad/s)  v_C(m/s)  alpha_BC(rad/s^2)  a_C(m/s^2)  a_C numerical
    0      -6.667     0.000              0.0      -53.33      -53.33
   30      -5.855    -1.293             61.8      -41.69      -41.69
   60      -3.482    -2.034            116.9      -13.35      -13.35
   90       0.000    -2.000            141.4       14.14       14.14
  120       3.482    -1.431            116.9       26.65       26.65
  150       5.855    -0.707             61.8       27.59       27.59
  180       6.667     0.000              0.0       26.67       26.67

The table shows something a designer needs to know. The slider's acceleration is largest in magnitude at 0°, the outer dead centre, where it is \(r\omega^2(1 + r/L) = 53.3 \text{ m/s}^2\), and smaller at 180°, the inner dead centre, where it is \(r\omega^2(1 - r/L) = 26.7 \text{ m/s}^2\). The asymmetry comes from the finite length of the rod. The force needed to accelerate the slider and whatever it pushes peaks at the outer dead centre, so that is where the connecting rod, its pins and the crank bearings see their highest inertia loads. Lesson 10 turns these accelerations into forces.

Motion relative to rotating axes

Sometimes the most natural description of a mechanism uses a reference frame that rotates: a pin sliding in a slot of a rotating arm, a block moving along a turning robot link, a quick-return mechanism. For a point \(B\) whose motion is described relative to axes attached to a body rotating at \(\boldsymbol{\Omega}\) about a point \(A\), the velocity and acceleration are

The subscript rel means as measured by an observer riding on the rotating axes. The first three acceleration terms are those of a point fixed to the rotating body; the last is the acceleration seen by the rotating observer; and the fourth, \(2\boldsymbol{\Omega}\times\mathbf{v}_{rel}\), is the Coriolis acceleration. For a point moving straight out along a rotating arm, these terms reduce exactly to the polar-coordinate result of Lesson 2, with the Coriolis term \(2\dot r\dot\theta\). The Coriolis acceleration is not a correction or an approximation; it is a real acceleration in the ground frame, and the slot or guide must supply the force to produce it.