Watch a cricket ball thrown in from the deep. Near the fielder's hand the path is almost a straight line, climbing steeply. At the top it turns over quickly. Near the keeper it is almost straight again. The path is curved everywhere, but not equally curved. How can "more curved" be given a number?
A circle for every point of a curve
Look at a small piece of any smooth curve under a magnifying glass. It looks like a piece of a circle. For every point \(P\) of the curve there is one circle that fits the curve best at \(P\). It touches the curve at \(P\), shares its tangent there, and bends at the same rate. This circle is the osculating circle, from the Latin for "kissing".
The radius of the osculating circle is the radius of curvature \(R\) of the curve at \(P\). Its centre, the centre of curvature, lies on the normal to the curve, on the inner side of the bend. A small \(R\) means a sharp bend. A large \(R\) means a gentle bend. A straight line has no bend at all, and its \(R\) is infinite. For a circle, the osculating circle is the circle itself, so \(R\) is its radius everywhere.
The radius of curvature from the motion
Now let a body move along the curve. Over a very short time near \(P\), the body moves as if it were on the osculating circle. On that circle its normal acceleration is the centripetal acceleration of Lesson 3, \(v^2/R\). So the normal acceleration of the body at \(P\) is \(a_{\perp} = v^2/R\). Turned round, this gives \(R\) from the speed and the acceleration.
The second form avoids finding \(a_{\perp}\) by hand. The size of \(\vec v \times \vec a\) is \(v\,a\sin\theta\), where \(\theta\) is the angle between the two vectors. Since \(a\sin\theta = a_{\perp}\), it equals \(v\,a_{\perp}\). Dividing \(v^3\) by it gives \(v^2/a_{\perp}\) again. In a plane, \(|\vec v \times \vec a| = |v_x a_y - v_y a_x|\), which is quick to work out from components.
The radius of curvature of a path at a point is the radius of the osculating circle there. That circle shares the tangent of the path and bends at the same rate. For a body moving along the path, it equals the square of the speed divided by the normal acceleration.
At one instant a particle has velocity \(\vec v = 3\hat i + 4\hat j\,\text{m/s}\) and acceleration \(\vec a = 2\hat i\,\text{m/s}^2\). Find its normal acceleration and the radius of curvature of its path at that point.
The speed is \(v = \sqrt{9 + 16} = 5\,\text{m/s}\).
\(|\vec v \times \vec a| = |v_x a_y - v_y a_x| = |3 \times 0 - 4 \times 2| = 8\,\text{m}^2/\text{s}^3\), so \(a_{\perp} = 8/5 = 1.6\,\text{m/s}^2\).
\(R = v^2/a_{\perp} = 25/1.6 \approx 15.6\,\text{m}\). Check with the other form: \(v^3/|\vec v \times \vec a| = 125/8 \approx 15.6\,\text{m}\).
A particle moves with velocity \(\vec v = 6\hat i\,\text{m/s}\) and acceleration \(\vec a = 2\hat i + 3\hat j\,\text{m/s}^2\). What is the radius of curvature of its path at this instant, in metres?
The curvature of a projectile's path
A projectile has one acceleration, \(\vec g\), straight down. At each point, split \(\vec g\) along the velocity and across it. If the velocity makes an angle \(\phi\) with the horizontal, the part across it is \(a_{\perp} = g\cos\phi\). The part along it, \(g\sin\phi\), only changes the speed. So \(R = v^2/(g\cos\phi)\) at any point of the path.
At the top of the path the velocity is horizontal. There \(\phi = 0\), so the whole of \(g\) acts across the velocity. The speed is the horizontal component, \(u\cos\theta\), for a launch at speed \(u\) and angle \(\theta\). At the launch point the speed is \(u\), but only \(g\cos\theta\) acts across the velocity.
The top is where the path is bent most sharply, for two reasons that work together. The speed is smallest there, and the whole of \(g\) acts across the velocity. Away from the top, the speed grows and a smaller part of \(g\) turns the velocity. Both make \(R\) larger. Since the horizontal velocity \(u\cos\theta = v\cos\phi\) never changes, the general result can be written \(R = v^3/(g\,u\cos\theta)\). It grows as the cube of the speed.
A ball is launched at \(20\,\text{m/s}\) at \(60°\) above the horizontal. Take \(g = 10\,\text{m/s}^2\). What is the radius of curvature of its path at the highest point, in metres?
A ball is thrown horizontally at \(10\,\text{m/s}\) from the roof of a tall building. Take \(g = 10\,\text{m/s}^2\). Find the radius of curvature of its path at the moment of release and \(1\,\text{s}\) later.
At release the velocity is horizontal and \(g\) is at right angles to it: \(R = v^2/g = 100/10 = 10\,\text{m}\). The release point is the top of this path.
After \(1\,\text{s}\): \(v_x = 10\,\text{m/s}\) and \(v_y = 10\,\text{m/s}\) downward, so \(v = 10\sqrt2 \approx 14.1\,\text{m/s}\) at \(45°\) below the horizontal.
The part of \(g\) across the velocity is \(g\cos 45° \approx 7.07\,\text{m/s}^2\). So \(R = 200/7.07 \approx 28.3\,\text{m}\). Check: \(v^3/(g v_x) = 2828/100 \approx 28.3\,\text{m}\).
Along the path of a projectile on level ground, where is the radius of curvature smallest?
Angular velocity about any point
Angular velocity was defined for motion on a circle about its centre. The idea works for any motion and any point. Choose a fixed point \(O\) and join it to the particle \(P\). This line is the line of sight. As \(P\) moves, the line of sight turns. The rate at which it turns is the angular velocity of \(P\) about \(O\).
Split the velocity of \(P\) into two parts. The part along the line of sight, \(v_{\parallel}\), only changes the distance \(OP\). It does not turn the line. The part at right angles to the line, \(v_{\perp}\), turns it, exactly as the velocity of a body on a circle of radius \(r = OP\) would. So \(\omega = v_{\perp}/r\).
The figure makes a surprising point. The particle moves in a straight line at a steady speed, yet it has an angular velocity about \(O\), and that angular velocity changes. For motion along a line at distance \(d\) from \(O\), \(v_{\perp} = v\,d/r\), so \(\omega = v\,d/r^2\). A different point \(O\) would give different values. The angular velocity of a single particle always needs the point to be named.
All points of a rigid body share one \(\omega\), and there the phrase "the angular velocity" is enough. For a single particle it is not. Its angular velocity about one point differs from its angular velocity about another, so always say about which point.
A car moves along a straight road at \(6\,\text{m/s}\). A watchman stands \(3\,\text{m}\) from the road. Find the angular velocity of the car about the watchman at the closest point. Find it again when his line of sight to the car makes \(30°\) with the road.
At the closest point the line of sight is at right angles to the road, so \(v_{\perp} = v = 6\,\text{m/s}\) and \(r = 3\,\text{m}\). \(\omega = 6/3 = 2\,\text{rad/s}\).
At \(30°\): the distance is \(r = 3/\sin 30° = 6\,\text{m}\), and \(v_{\perp} = v\sin 30° = 3\,\text{m/s}\). So \(\omega = 3/6 = 0.5\,\text{rad/s}\).
The watchman's head turns four times as fast at the closest point as at the \(30°\) point. That is why a passing car seems to "whip" past.
A particle is at \((3, 4)\,\text{m}\) and moves with velocity \(2\hat i\,\text{m/s}\). What is its angular velocity about the origin?
Christiaan Huygens had studied how sharply a curve bends in his work on pendulum clocks. In 1686 Gottfried Wilhelm Leibniz, the co-inventor of calculus, gave the best-fitting circle its name. He called it the circulus osculans, the kissing circle, because it touches the curve more closely than any other circle can. The name "osculating circle" is still used, and its radius is the radius of curvature of this lesson.
Suppose a train ran from a straight track directly into a circular curve. Its normal acceleration would jump from zero to \(v^2/R\) in an instant, a sharp jerk for the passengers. So railway engineers join a straight track to a curve by a transition curve, whose radius of curvature falls smoothly from infinity to \(R\). Along it, \(a_{\perp} = v^2/R\) rises gradually. Highways and the ramps of flyovers use the same idea.
- Using the whole acceleration in \(R = v^2/a\).Only the part across the velocity bends the path. Use \(a_{\perp}\), or \(v^3/|\vec v \times \vec a|\).
- Taking the radius of curvature at the launch of a projectile to be \(u^2/g\).At the launch only \(g\cos\theta\) is across the velocity: \(R = u^2/(g\cos\theta)\). \(u^2/g\) holds only for a horizontal launch.
- Saying that a particle moving in a straight line has no angular velocity.About a point off the line it has \(\omega = v_{\perp}/r\). It is zero only about points on the line of motion.
- Quoting a particle's angular velocity without naming the point.For a single particle, \(\omega\) depends on the point. Always say about which point.