After school, a boy swings his bag by its strap in a big vertical circle. His hand feels a hard tug each time the bag sweeps past the bottom. At the top the strap goes almost loose, and the tug nearly vanishes. The bag is the same, and the strap is the same. So what changes from the bottom to the top?
Why a vertical circle is different
In the chapter on Dynamics of Circular Motion, bodies went round at a steady speed, like a stone on a smooth table. In a vertical circle, the weight lies in the plane of the motion. As the body rises, part of its weight acts against the motion and slows it. As the body falls, the same part speeds it up. So the speed changes from point to point, and the pull that the string must supply changes too.
One angle for the position
Take a small bob of mass \(m\) tied to a light string of length \(r\). The other end of the string is fixed at the point O. We give the position of the bob by the angle \(\theta\) that the string makes with the downward vertical. So \(\theta = 0\) is the lowest point, \(\theta = 90°\) is level with O, and \(\theta = 180°\) is the top. The height of the bob above the lowest point is \(h = r(1 - \cos\theta)\), which is \(2r\) at the top.
Two forces, split two ways
Two forces act on the bob: the tension \(T\) along the string towards O, and the weight \(mg\) straight down. We split each force along two directions at the bob. One direction is the radius, along the string. The other is the tangent, along the path. The tension lies wholly along the radius. The weight makes the angle \(\theta\) with the line of the string. So it splits into \(mg\cos\theta\) along the string, pointing away from O, and \(mg\sin\theta\) along the tangent, pointing back towards the bottom.
Each part has its own job
The two directions do different jobs, as the chapter on Kinematics of Circular Motion showed. The parts along the radius turn the velocity. Their net inward sum must equal \(mv^2/r\), where \(v\) is the speed at that instant. The parts along the tangent change the speed. Here only \(mg\sin\theta\) acts along the tangent, so the tangential acceleration is \(g\sin\theta\), directed towards the bottom.
\[ T - mg\cos\theta = \frac{mv^2}{r}, \qquad a_t = g\sin\theta \]
The tangential acceleration explains the changing speed. While the bob climbs, from \(\theta = 0\) to \(180°\), the acceleration \(g\sin\theta\) points back down the path, so the bob slows. On the way down it points along the motion, so the bob speeds up. At the very top and the very bottom, \(\sin\theta = 0\), and the speed is momentarily not changing.
There are only two forces on the bob: \(T\) and \(mg\). Do not draw a third arrow called the centripetal force. "Centripetal" is the name of the job that the net radial force does, here \(T - mg\cos\theta\). Adding it as an extra force counts the same force twice.
The bob is level with O, so the string is horizontal. Which force supplies the whole centripetal requirement at this instant?
Energy gives the speed
The tangential equation could be solved step by step, but energy gives the speed in one line. The tension always points along the radius, at right angles to the velocity, so it does no work. Only gravity does work, and gravity is a conservative force. So the mechanical energy of the bob stays the same all the way round, as in the chapter on Potential Energy and Conservation of Energy.
Let \(u\) be the speed at the lowest point, and take the zero of potential energy there. At the angle \(\theta\) the bob is higher by \(r(1 - \cos\theta)\). Equate the energy there to the energy at the bottom:
\[ \tfrac12 mu^2 = \tfrac12 mv^2 + mgr(1 - \cos\theta) \]
The speed depends only on the height. The bob has the same speed at \(\theta = 60°\) on the way up and on the way down. At the level of O, the formula gives \(v^2 = u^2 - 2gr\). At the top, it gives \(v^2 = u^2 - 4gr\), the smallest value round the circle.
A \(0.2\,\text{kg}\) bob on a string \(0.5\,\text{m}\) long moves at \(5\,\text{m/s}\) at the lowest point. Take \(g = 10\,\text{m/s}^2\). What is its speed, in m/s, when the string is horizontal?
The tension follows
Now return to the radial equation and solve it for the tension. Energy has already given \(v^2\) at every angle, so the tension at every angle follows by substitution.
\[ T - mg\cos\theta = \dfrac{mv^2}{r} \]
Why. along the string, the net force towards O must equal \(mv^2/r\). The tension points to O and the part \(mg\cos\theta\) points away.
\[ T = \dfrac{mv^2}{r} + mg\cos\theta \]
Why. move the weight's part to the other side.
\[ T = \dfrac{m}{r}\left[u^2 - 2gr(1 - \cos\theta)\right] + mg\cos\theta \]
Why. put in the speed from energy conservation.
\[ T = \dfrac{mu^2}{r} - 2mg + 3mg\cos\theta \]
Read it. open the bracket; \(2mg\cos\theta\) and \(mg\cos\theta\) add to \(3mg\cos\theta\).
The factor \(\cos\theta\) falls from \(1\) at the bottom to \(-1\) at the top, so the tension falls steadily as the bob climbs. Two causes work together. At the bottom the bob is fastest, and its weight points away from O, so the string must supply the weight as well as \(mv^2/r\). At the top the bob is slowest, and its weight points towards O, so gravity does part of the string's job.
Read the three points of Fig 1.2 from the formula, with \(m = 0.5\,\text{kg}\), \(r = 1\,\text{m}\) and \(u = 8\,\text{m/s}\). At the bottom, \(T = mu^2/r + mg = 32 + 5 = 37\,\text{N}\). Level with O, \(\cos\theta = 0\), so \(T = mv^2/r\) with \(v^2 = 44\), which gives \(22\,\text{N}\). At the top, \(v^2 = 24\) and \(T = mv^2/r - mg = 12 - 5 = 7\,\text{N}\).
A \(0.5\,\text{kg}\) ball on a string \(1\,\text{m}\) long moves at \(6\,\text{m/s}\) at the lowest point of a vertical circle. Take \(g = 10\,\text{m/s}^2\). Find the speed and the tension (a) at the lowest point, (b) at \(\theta = 60°\), (c) when the string is horizontal. (d) Can the ball reach the top on the circle?
(a) At the bottom \(v = 6\,\text{m/s}\) and \(T = mu^2/r + mg = 0.5 \times 36 + 5 = 23\,\text{N}\).
(b) \(v^2 = 36 - 20(1 - 0.5) = 26\), so \(v \approx 5.10\,\text{m/s}\). Then \(T = mv^2/r + mg\cos 60° = 13 + 2.5 = 15.5\,\text{N}\).
(c) \(v^2 = 36 - 20 = 16\), so \(v = 4\,\text{m/s}\). The weight has no part along the string, so \(T = mv^2/r = 0.5 \times 16 = 8\,\text{N}\).
(d) At the top the formula would give \(v^2 = 36 - 40 = -4\), which is impossible. The ball cannot reach the top on the circle. Lesson 3 finds where it leaves the circle.
A bob goes all the way round a vertical circle on a string. Rank these tensions from the largest to the smallest.
A \(0.4\,\text{kg}\) bob is held with its string horizontal and released from rest. Take \(g = 10\,\text{m/s}^2\). What is the tension, in newtons, as it passes the lowest point?
A bob goes round a vertical circle on a string. Which statements are true?
In 1659 the Dutch physicist Christiaan Huygens wrote a short work on the force of a body moving in a circle. He was designing pendulum clocks, and he wanted to know how hard a swinging bob pulls on its thread. One of his results was striking. A bob released with its thread horizontal pulls, at the lowest point, with three times its own weight. The length of the thread does not matter. The check above gives the same answer, \(3mg\), from energy and the radial equation.
Back to the boy and his bag. At the bottom of each circle the strap must hold up the weight of the bag and also supply \(mv^2/r\) at the greatest speed. At the top the weight of the bag does part of the job, and the bag moves slowest. So the tug on his hand is largest at the bottom and smallest at the top. If he slows down, the strap at the top goes loose, and the bag drops out of its circle. The next lesson finds how slow is too slow.
- Taking the tension as \(mv^2/r\) at every point.That holds only level with O. Elsewhere the weight has a radial part, so \(T = mv^2/r + mg\cos\theta\).
- Using the same speed all the way round, as in uniform circular motion.The weight has a part along the tangent, so the speed changes. Find it from energy, \(v^2 = u^2 - 2gr(1 - \cos\theta)\).
- Measuring \(\theta\) from the top or from the horizontal and then using the formula of this lesson.In this chapter \(\theta\) is measured at the centre from the lowest point. Check the formula at \(\theta = 0\), where it must give the bottom values.
- Drawing a separate centripetal force on the bob.Only \(T\) and \(mg\) act. Their net radial part is what does the centripetal job.