A painter in Jaipur leans a ladder against a freshly painted wall. The foot slips on the smooth floor, and the ladder slides down the wall. A door turns about its hinges, and a fan turns about its axle. The ladder has no hinge and no axle. Yet it clearly turns as it falls. About what does it turn?
A point that does not move
Lesson 1 gave the velocity of any point as a translation part plus a rotation part. Could the two parts cancel somewhere? The rotation part at P has size \(\omega r\) and points at right angles to OP. So the right point, at the right distance from O, has a rotation part exactly equal and opposite to \(\vec v_O\). That point is at rest.
Use the plate of lesson 1. Its point O moves at \(1.2\,\text{m/s}\) along \(+x\), and the plate turns anticlockwise at \(2\,\text{rad/s}\). A point above O moves to the left relative to O, since the plate turns anticlockwise. Its rotation part is \(\omega d\), where \(d\) is its distance above O. The two parts cancel when \(\omega d = v_O\), that is when \(d = 1.2/2 = 0.6\,\text{m}\).
Call that point I. The line through I at right angles to the plane is the instantaneous axis of rotation. Now use I as the reference point. Its velocity is zero, so the translation part vanishes, and the rule of lesson 1 becomes \(\vec v_P = \vec\omega\times\vec r_{P/I}\). That is a pure rotation about I. At this instant the whole plate moves as if it were pinned at I and turning about it.
The instantaneous axis of rotation of a body in plane motion passes through the point I that is at rest at that instant. It is at right angles to the plane. At that instant every point of the body moves as in a pure rotation about I.
Fig 2.1 checks the numbers of lesson 1 in one picture. O is \(0.6\,\text{m}\) from I, so it moves at \(2 \times 0.6 = 1.2\,\text{m/s}\). P is \(\sqrt{0.8^2 + 0.6^2} = 1.0\,\text{m}\) from I, so it moves at \(2.0\,\text{m/s}\), at right angles to IP. Q is also \(1.0\,\text{m}\) from I and also moves at \(2.0\,\text{m/s}\). Lesson 1 needed two parts at each point. The axis needs one distance.
Notice two things about I. First, it lies outside the plate. The axis is a place, not a piece of the body; it may fall on the body or beside it. Second, a body in pure translation has \(\omega = 0\), so \(d_I = v_O/\omega\) has no finite value. Its axis is infinitely far away, which is another way of saying that every point moves alike.
The skidding wheel of Fig 1.2 has radius \(0.5\,\text{m}\). Its centre moves at \(4\,\text{m/s}\) and it turns in the rolling sense at \(4\,\text{rad/s}\). How far below the road is its instantaneous axis, in metres?
Finding the axis
Every velocity is at right angles to the line joining its point to I. Read that fact the other way. If you know the direction of the velocity of a point, the axis lies somewhere on the line through that point perpendicular to its velocity. Two such lines, from two points, meet at the axis.
There are three common ways to use this.
- Two velocity directions are known. Draw the perpendicular to each velocity through its own point. They meet at I.
- A point at rest is known. Then it lies on the axis. The point of a wheel that touches the road without sliding is the standard case.
- Two velocities are parallel. Then both perpendiculars are one line, and I lies on it. The speeds are in proportion to the distances from I, which fixes the place.
In the third case, compare the directions. If the two parallel velocities point the same way, the axis lies beyond the slower point. If they point opposite ways, it lies between the two points. If they are equal and point the same way, there is no axis at all: the body translates.
A ladder \(5\,\text{m}\) long slides with its foot A on a smooth floor and its top B against a smooth wall. At one instant A is \(4\,\text{m}\) from the wall and moves away from it at \(1.5\,\text{m/s}\). Find the angular velocity of the ladder, the speed of B and the speed of the midpoint M.
Put the corner at the origin. A is at \((4, 0)\), and B is \(\sqrt{5^2 - 4^2} = 3\,\text{m}\) up the wall, at \((0, 3)\). The velocity of A is horizontal, and the velocity of B is vertical.
The perpendicular to \(\vec v_A\) through A is the vertical line \(x = 4\). The perpendicular to \(\vec v_B\) through B is the horizontal line \(y = 3\). They meet at I \((4, 3)\).
A is \(3\,\text{m}\) below I, so \(\omega = v_A/3 = 0.5\,\text{rad/s}\). B is \(4\,\text{m}\) from I, so \(v_B = 0.5 \times 4 = 2.0\,\text{m/s}\), down the wall. M is at \((2, 1.5)\), a distance \(\sqrt{2^2 + 1.5^2} = 2.5\,\text{m}\) from I, so \(v_M = 0.5 \times 2.5 = 1.25\,\text{m/s}\).
The ladder turns at \(0.5\,\text{rad/s}\), B slides down at \(2.0\,\text{m/s}\) and M moves at \(1.25\,\text{m/s}\) at right angles to IM. The rule of equal components along the ladder agrees: \(1.5 \times 0.8 = 2.0 \times 0.6 = 1.2\,\text{m/s}\).
The chapter on Rigid Bodies found the speed of the top of such a ladder by equal components along it. The axis gives more for the same effort. One picture gives the angular velocity, the speed of the top and the speed of any other point.
Watch I in Fig 2.2. It always sits straight above A and level with B, at \((x_A, y_B)\). Its distance from the corner is \(\sqrt{x_A^2 + y_B^2}\), and that is the length of the ladder, \(5\,\text{m}\), at every instant. So the axis moves on a quarter circle about the corner. Pause the figure when A is \(4\,\text{m}\) out: the readouts give \(\omega = 0.50\,\text{rad/s}\) and \(v_B = 2.00\,\text{m/s}\), as in the example.
Notice also how \(\omega\) grows as the ladder flattens, from \(0.31\) to \(0.69\,\text{rad/s}\). The foot keeps a steady speed, but its distance below I, which is the height of B, shrinks. So \(\omega = v_A/y_B\) rises, and B falls faster and faster.
Two points P and Q of a rigid plate lie on a vertical line, with P \(0.5\,\text{m}\) above Q. At one instant P moves at \(3\,\text{m/s}\) and Q at \(1\,\text{m/s}\), both to the right. What is the angular velocity of the plate, in rad/s?
Two points of a rigid body lie on a vertical line, \(0.4\,\text{m}\) apart. The upper point moves at \(3\,\text{m/s}\) to the right and the lower point at \(1\,\text{m/s}\) to the left. Where is the instantaneous axis?
A cat sits at the middle of a ladder of length \(L\) that slides down a smooth wall on a smooth floor. What path does the cat follow?
An axis for one instant
The word instantaneous matters. At each instant the velocities fix one axis. A moment later the velocities have changed, and so has the axis. For the ladder it travels along a quarter circle. It is not a fixed point of space, and it is not a fixed particle of the body either.
A particle that lies at the axis has zero velocity at that instant. It does not have zero acceleration. If it did, it would stay at rest, and it would remain on the axis. But the axis moves on, and a different particle takes its place. The particle that left must have gained a velocity, so it was accelerating.
The sharpest case is a wheel that rolls on a road without sliding, the subject of the next lesson. The point of the tyre touching the road is at rest. Yet it rises from the road an instant later, so it accelerates, and lesson 4 finds that acceleration. A point with zero velocity can still have a large acceleration.
The axis answers questions about velocities: the speed of a point, its direction, and \(\omega\). It does not answer questions about accelerations, because accelerations about I are not those of a fixed axle. Do not write \(a = \alpha d\) about I, and do not use I as a fixed pivot in a torque equation. The chapter on Dynamics of Rolling says when that shortcut is allowed.
A ladder slides down a smooth wall. Which statement about its instantaneous axis is correct?
A particle of a body lies at the instantaneous axis. Explain why its acceleration need not be zero.
The particle has zero velocity only at that instant. The axis moves, and a moment later other particles lie on it, so this particle is no longer at rest. Its velocity has changed from zero to some value, which means it has an acceleration. Zero velocity at one instant says nothing about the rate at which the velocity is changing.
In 1763 the Florentine mathematician Giulio Mozzi published a short treatise on the motion of rigid bodies. He showed that at any instant a moving body turns about one line, and at most slides along that line as well. He called it the instantaneous axis. Nearly seventy years later Michel Chasles proved the same result for finite motions. Mozzi's axis is the one this lesson uses, in its plane form.
Return to the painter's ladder. It has no hinge, but at every instant it turns about the point above its foot and level with its top. That point is the instantaneous axis. It travels round a quarter circle as the ladder slides, and the ladder turns faster about it as the top comes down.
- Treating the instantaneous axis as a fixed pivot.Find it again at every instant. It moves through space and from particle to particle.
- Drawing the perpendicular to a velocity through the wrong point.The perpendicular must pass through the point whose velocity it is. Draw the point, its velocity, then the line at right angles through that point.
- Giving the particle at the axis zero acceleration.Only its velocity is zero. Use the axis for velocities and \(\omega\), not for accelerations.