Stand beside a road in Kolkata and watch a cycle-rickshaw pass. Look hard at one of its wheels. The spokes near the top are a grey blur. The spokes near the road are sharp, and you can almost count them. Every spoke belongs to the same wheel and turns at the same rate. So why does the wheel look fast at the top and slow at the bottom?
From a picture to a number
The chapter on Rigid Bodies and Moment of Inertia described any motion of a rigid body in two parts. One point translates, and the body rotates about that point. That was a picture. This lesson turns the picture into a rule that gives the velocity of every point of the body, one point at a time.
Choose any point O of the body and call it the reference point. Imagine riding with O in a frame that moves with O but does not turn. In that frame O is at rest. A rigid body keeps every distance fixed, so every other point P must stay at the same distance from O. The only motion left to P is a circle round O.
A point moving on a circle of radius \(r\) at angular velocity \(\omega\) has speed \(\omega r\), directed along the circle. So, relative to O, the point P moves at \(\omega r\) at right angles to the line OP. Here \(r\) is the distance OP. Now return to the ground. The rule of relative velocity from Relative Velocity in Two Dimensions adds the velocity of the frame to the velocity seen in it.
\[ \vec v_P = \vec v_O + \vec v_{P/O} \]
Why. the rule of relative velocity: the velocity of P on the ground is the velocity of O plus the velocity of P relative to O.
\[ |\vec v_{P/O}| = \omega r, \quad \vec v_{P/O} \perp \vec r_{P/O} \]
Why. relative to O, the point P moves on a circle of radius \(r = OP\), because the body is rigid.
\[ \vec v_{P/O} = \vec\omega\times\vec r_{P/O} \]
Why. the cross product has size \(\omega r\) and points at right angles to OP, in the sense of the turning, by the right-hand rule.
\[ \vec v_P = \vec v_O + \vec\omega\times\vec r_{P/O} \]
Read it. put step 3 into step 1.
For any two points O and P of a rigid body moving in a plane, \(\vec v_P = \vec v_O + \vec\omega\times\vec r_{P/O}\). The first term is the translation part, the same for every point. The second is the rotation part, of size \(\omega \cdot OP\), at right angles to OP.
In a plane you rarely need the cross product itself. Two facts are enough. The rotation part has size \(\omega r\), and it points at right angles to OP, in the direction the body is turning. Draw the translation arrow at P, draw the rotation arrow from its head, and join the tail to the final head.
A flat plate moves on a smooth floor. Its point O moves at \(1.2\,\text{m/s}\) along \(+x\), and the plate turns anticlockwise at \(2\,\text{rad/s}\). The point P lies \(0.8\,\text{m}\) from O along \(+x\). Find the velocity of P.
Take O as the reference point. The translation part is \(1.2\,\text{m/s}\) along \(+x\). The rotation part has size \(\omega r = 2 \times 0.8 = 1.6\,\text{m/s}\).
OP lies along \(+x\). A point to the right of O, on a plate turning anticlockwise, moves straight up relative to O. So the rotation part is \(1.6\,\text{m/s}\) along \(+y\).
\(\vec v_P = (1.2, 1.6)\,\text{m/s}\). The two parts are at right angles, so \(v_P = \sqrt{1.2^2 + 1.6^2} = 2.0\,\text{m/s}\).
P moves at \(2.0\,\text{m/s}\) at \(\tan^{-1}(1.6/1.2) \approx 53°\) above the \(+x\) direction. It is faster than O because its rotation part adds a sideways velocity.
A rod \(1\,\text{m}\) long lies along the x axis on a smooth floor. Its centre C moves at \(1.5\,\text{m/s}\) along \(+y\), and the rod turns anticlockwise at \(4\,\text{rad/s}\). The end A is at the left, \(0.5\,\text{m}\) from C. What is the speed of A, in m/s?
Which point should be O?
The rule leaves a choice open. Any point of the body may serve as the reference point. It seems possible that two choices could give two different angular velocities for the same body. They cannot, and the proof needs only the rule itself.
\[ \vec v_P = \vec v_O + \vec\omega\times\vec r_{P/O} \]
Why. the rule about O, applied to the point P.
\[ \vec v_Q = \vec v_O + \vec\omega\times\vec r_{Q/O} \]
Why. the same rule about O, applied to another point Q.
\[ \vec v_P - \vec v_Q = \vec\omega\times(\vec r_{P/O} - \vec r_{Q/O}) \]
Why. subtract step 2 from step 1; the velocity of O cancels.
\[ \vec v_P = \vec v_Q + \vec\omega\times\vec r_{P/Q} \]
Read it. \(\vec r_{P/O} - \vec r_{Q/O}\) is the position of P measured from Q. This is the rule about Q, with the same \(\vec\omega\).
So the rule about Q uses exactly the same \(\vec\omega\) as the rule about O. The angular velocity belongs to the body, not to the point we describe it from. What does change is the split. A different reference point has a different velocity, so the translation part changes, and the rotation part changes with it. Their sum at P never changes.
Switch the chip in Fig 1.1 to "about Q". The point Q is \(0.4\,\text{m}\) below O, and its own velocity is \(2.0\,\text{m/s}\) along \(+x\). About Q, the translation part at P is \(2.0\,\text{m/s}\), and the rotation part is \(2 \times 0.89 = 1.79\,\text{m/s}\) at right angles to QP. The grey arrows are now different, yet the blue arrow at P is the same \(2.0\,\text{m/s}\) at \(53°\).
Paint a straight line on the plate. In one second it turns through the same angle, whichever point of the plate you follow with your eye. Following a different point only changes how the line drifts across the floor, never how fast it turns. That is why the body has one \(\omega\).
In practice we choose the point that makes the work easiest. Usually it is the centre of mass, because its motion follows from the external forces alone, as the chapter on Centre of Mass showed. For a wheel on a road, a point of the rim touching the road is sometimes easier still. The next two lessons show why.
Two students study the same tumbling plate at the same instant. One takes its centre of mass as the reference point. The other takes a corner. Which statement is correct?
Explain why a rigid body has only one angular velocity, whatever point is used to describe its motion.
Write the rule about O for two points P and Q, and subtract. The velocity of O cancels, and what remains is the rule about Q with the same angular velocity. Physically, a line painted on the body turns through one angle in a given time, whichever point of the body is followed. Only the translation part depends on the point chosen.
A wheel that moves and turns
Now apply the rule to a wheel of radius \(R\). Take its centre as the reference point. The centre moves along the road at \(v_{cm}\), and the wheel turns at \(\omega\). A wheel moving to the right on a road usually turns clockwise, as seen with the road below and the wheel's motion to the right. We call that the rolling sense of turning.
Every point of the rim is at distance \(R\) from the centre, so every rim point has a rotation part of the same size, \(\omega R\). Only its direction changes round the rim. At the top it points forward. At the bottom it points backward. At the front of the wheel it points down, and at the back it points up.
Add the translation part \(v_{cm}\), forward at every point. At the top the two parts point the same way, and at the bottom they oppose. At the front and the back they are at right angles.
\[ v_{\text{top}} = v_{cm} + \omega R, \qquad v_{\text{bottom}} = v_{cm} - \omega R, \qquad v_{\text{front}} = v_{\text{back}} = \sqrt{v_{cm}^2 + \omega^2R^2} \]
The bottom value may be positive, zero or negative. A negative value means the bottom point moves backward over the road.
In Fig 1.2 the rotation part is \(\omega R = 4 \times 0.5 = 2\,\text{m/s}\) at every rim point. At the top, \(4 + 2 = 6\,\text{m/s}\). At the bottom, \(4 - 2 = 2\,\text{m/s}\), still forward. At the front, \(\sqrt{4^2 + 2^2} = \sqrt{20} \approx 4.47\,\text{m/s}\), tilted \(\tan^{-1}(2/4) \approx 27°\) below the road. The back point moves at the same speed, tilted \(27°\) above it.
Consider the bottom point of this wheel. It touches the road, yet it moves forward over the road at \(2\,\text{m/s}\). The wheel is sliding, as a tyre does on a wet road when the brakes are pressed. A wheel whose bottom point is at rest is special, and lesson 3 is about it.
The rim of a turning wheel has one \(\omega\) and one value of \(\omega R\), but not one speed. The rotation part has the same size at every rim point, and its direction changes round the rim. Added to the same translation part, it gives a different sum at each point.
The wheel of a toy car, of radius \(0.2\,\text{m}\), skids on a polished floor. Its centre moves at \(3\,\text{m/s}\), and it turns in the rolling sense at \(10\,\text{rad/s}\). What is the speed of the top point of its rim, in m/s?
A wheel of radius \(R\) moves to the right with its centre at \(v_{cm}\) and turns in the rolling sense, with \(\omega R\) larger than \(v_{cm}\). How does the point touching the road move?
A wheel moves to the right and turns in the rolling sense, with \(\omega R\) smaller than \(v_{cm}\). Rank these points of the wheel by their speed, fastest first.
The main rotor of a helicopter such as the Dhruv, built by HAL in Bengaluru, turns so that its blade tips move at about \(200\,\text{m/s}\). Now let the helicopter fly forward at \(70\,\text{m/s}\). On one side of the rotor a tip moves forward through the air at about \(270\,\text{m/s}\). On the other side a tip moves backward at only about \(130\,\text{m/s}\). The slow side gives less lift, and near the top speed the blades on that side can stall. The rule \(\vec v_P = \vec v_O + \vec\omega\times\vec r\) sets that limit.
Return to the cycle-rickshaw wheel at the start. Each point of the wheel carries the forward motion of the axle and the turning about the axle. At the top the two add, so the top spokes move fastest and blur. At the bottom they nearly cancel, so the lowest spokes move very slowly and look sharp. Lesson 3 shows that for a wheel that does not slide, they cancel exactly.
- Adding the rotation part along the direction of motion at every point.The rotation part is at right angles to the line from the reference point. Draw that line first, then the arrow across it.
- Giving a body different angular velocities about different points.A rigid body has one \(\omega\). Only the translation part depends on the reference point chosen.
- Forgetting the translation part at a rim point.Every point carries the full velocity of the reference point. The rotation part is added to it, never used in its place.