Children on a village swing, a jhula hung from a branch of a banyan tree, try to swing higher and higher. Most swings stay smooth. But when a swing climbs above the level of the branch, the ropes suddenly go loose. The child drops for a moment, and the ropes snap tight again with a jolt.
Which reaches zero first
Two quantities fall as the bob climbs: its speed and the tension. Whichever reaches zero first decides what happens next. From Lesson 1, \(v^2 = u^2 - 2gr(1 - \cos\theta)\) and \(T = mv^2/r + mg\cos\theta\). The answer depends on whether the bob is below or above the level of O.
Below the level of O, \(\cos\theta\) is positive. Then \(T = mv^2/r + mg\cos\theta\) is a sum of two positive terms. Even where the bob stops, the tension is still \(mg\cos\theta\), which is more than zero. So below the level of O the speed reaches zero first, and the string stays tight. The bob swings back like a pendulum.
Above the level of O, \(\cos\theta\) is negative, and the weight's radial part points towards O. It does some of the string's job, so \(T = mv^2/r - mg|\cos\theta|\) can fall to zero while \(v\) is still positive. So above the level of O the tension reaches zero first. The string goes slack while the bob is still moving. On a string, a bob can never come to rest on the circle above the level of O.
Three ranges of launch speed
The bob reaches the level of O only if \(u^2 \ge 2gr\), and it reaches the top on the circle only if \(u^2 \ge 5gr\). So three ranges of the speed \(u\) at the bottom give three kinds of motion:
- \(u \le \sqrt{2gr}\): the bob rises no higher than the level of O and swings back and forth, with the string tight all the time.
- \(\sqrt{2gr} < u < \sqrt{5gr}\): the bob rises above the level of O, and the string goes slack somewhere between \(90°\) and \(180°\).
- \(u \ge \sqrt{5gr}\): the bob goes all the way round, with the string tight everywhere.
Try the two ends of the middle range with the slider. At \(4.5\,\text{m/s}\) the string goes slack almost level with O, and the bob barely moves when it does. At \(7.0\,\text{m/s}\), only \(0.07\,\text{m/s}\) short of \(\sqrt{5gr}\), the string goes slack at \(165°\), close to the top. There is no sudden jump between the ranges. The slack point climbs smoothly from the level of O to the top as the launch speed grows.
A bob on a string \(1\,\text{m}\) long is given \(3\,\text{m/s}\) at the lowest point. Take \(g = 10\,\text{m/s}^2\). What happens?
Finding the slack point
The string goes slack at the angle \(\theta_s\) where the tension first becomes zero. Set \(T = 0\) in the formula of Lesson 1 and solve for the angle. Then energy gives the height of the point and the speed of the bob there.
\[ \dfrac{mu^2}{r} - 2mg + 3mg\cos\theta_s = 0 \]
Why. the tension formula of Lesson 1, set equal to zero at the slack point.
\[ 3gr\cos\theta_s = 2gr - u^2 \]
Why. multiply by \(r/m\) and move the other terms across.
\[ \cos\theta_s = -\dfrac{u^2 - 2gr}{3gr} \]
Why. divide by \(3gr\). For \(2gr < u^2 < 5gr\) the right side lies between \(0\) and \(-1\), so \(\theta_s\) lies between \(90°\) and \(180°\).
\[ h_s = r(1 - \cos\theta_s) = \dfrac{u^2 + gr}{3g} \]
Why. the height above the lowest point, after putting in \(\cos\theta_s\).
\[ v_s^2 = -gr\cos\theta_s = \dfrac{u^2 - 2gr}{3} \]
Read it. at the slack point \(T = 0\), so the radial equation leaves \(mv_s^2/r = -mg\cos\theta_s\), the weight's radial part.
The speed at the slack point is not zero. There the weight's radial part, \(-mg\cos\theta_s\), is exactly the force the circle needs, so the bob is in free fall for an instant. A moment later the bob is slower and higher, and the weight's radial part is more than the circle needs. To keep the bob on the circle, the string would have to push outwards. It cannot, so the bob leaves the circle, still moving.
For a quick case, take \(u^2 = 3gr\). The formula gives \(\cos\theta_s = -1/3\), so \(\theta_s \approx 109.5°\), and \(h_s = 4r/3\). The bob leaves the circle a third of the radius above the level of O, with \(v_s^2 = gr/3\).
A bob on a string \(1\,\text{m}\) long is launched at \(6\,\text{m/s}\) from the lowest point. Take \(g = 10\,\text{m/s}^2\). At what height, in metres above the lowest point, does the string go slack?
At the instant the string goes slack, what is the speed of the bob?
A bob on a string \(1\,\text{m}\) long is launched at \(7\,\text{m/s}\), just under \(\sqrt{5gr} \approx 7.07\,\text{m/s}\). Take \(g = 10\,\text{m/s}^2\). At what height, in metres above the lowest point, does the string go slack?
A projectile inside the circle
Once the string is slack, the only force on the bob is its weight. So the bob is a projectile, exactly as in the chapter on Motion in Two Dimensions. It starts from the slack point S with the speed \(v_s\), along the tangent to the circle. The bob is moving anticlockwise there, above the level of O, so its velocity points up and towards the other side. It makes the angle \(180° - \theta_s\) with the horizontal.
The weight pulls inwards harder than the circle needs, so the path bends more sharply than the circle. The bob falls inside the circle, and the string stays slack. The string tightens again only when the bob comes back to a distance \(r\) from O. At that instant the string jerks, and the bob's motion along the string stops suddenly. Some energy is lost in the jerk, so the energy methods of this chapter do not carry across it.
A bob on a string \(1.4\,\text{m}\) long is launched at \(7\,\text{m/s}\) from the lowest point. Take \(g = 10\,\text{m/s}^2\). Find (a) where the string goes slack, (b) the velocity there, (c) the highest point of the bob, and (d) where the string tightens again.
(a) \(gr = 14\), so \(\cos\theta_s = -(49 - 28)/42 = -0.5\) and \(\theta_s = 120°\). The height is \(h_s = 1.4 \times 1.5 = 2.1\,\text{m}\) above the lowest point, and S is \(1.4\sin 120° \approx 1.21\,\text{m}\) to the side of O.
(b) \(v_s^2 = -gr\cos\theta_s = 7\), so \(v_s \approx 2.65\,\text{m/s}\), at \(180° - 120° = 60°\) above the horizontal, towards the other side. Its parts are \(2.65\cos 60° \approx 1.323\,\text{m/s}\) sideways and \(2.65\sin 60° \approx 2.291\,\text{m/s}\) up.
(c) The upward part carries the bob a further \(2.291^2/20 \approx 0.26\,\text{m}\), exactly \(7 \times 0.75/20 = 0.2625\,\text{m}\). The highest point is \(2.36\,\text{m}\) above the lowest point, \(0.44\,\text{m}\) below the top of the circle.
(d) The sideways speed takes \(1.212/1.323 \approx 0.9165\,\text{s}\) to bring the bob to the vertical through O. In that time its height changes by \(2.291 \times 0.9165 - 5 \times 0.9165^2 = 2.10 - 4.20 = -2.10\,\text{m}\). S was \(0.7\,\text{m}\) above O, so the bob arrives \(1.4\,\text{m}\) below O, at the lowest point of the circle. The string tightens there.
For the bob of the earlier check, on a string \(1\,\text{m}\) long launched at \(6\,\text{m/s}\), the string goes slack with \(\cos\theta_s = -8/15\) and \(v_s^2 = 16/3\,\text{m}^2/\text{s}^2\). Take \(g = 10\,\text{m/s}^2\). How far, in metres, does the bob rise above the slack point?
For one launch speed the parabola passes exactly through the centre O. The condition reduces to \(u^2 = (2 + \sqrt3)gr \approx 3.73\,gr\), with the string going slack at \(\cos\theta_s = -1/\sqrt3\), about \(125°\). Problems that ask for such a speed are solved like the example above. Write the projectile's position from S, and demand that it pass through the chosen point.
A swing is a bob on two ropes. A child pumping the swing adds energy each time, so the speed at the bottom grows. While the swing stays below the level of the branch, \(u \le \sqrt{2gr}\), and the ropes stay tight. Once it climbs above the branch, the ropes go slack before the swing stops. The child falls freely for a moment, on a parabola inside the circle. Then the ropes snap tight, and the jolt is the jerk of the string. With ropes \(3\,\text{m}\) long, the swing goes above the branch once its speed at the bottom passes \(\sqrt{60} \approx 7.7\,\text{m/s}\).
- Setting \(v = 0\) to find where the bob leaves the circle.Above the level of O the tension reaches zero first. Set \(T = 0\), not \(v = 0\).
- Thinking the string can go slack below the level of O.Below O, \(T = mv^2/r + mg\cos\theta\) is positive even at rest. The bob only swings.
- Treating the motion after the slack point as circular.With the string slack only gravity acts, so the bob follows a parabola that starts along the tangent at S.
- Using energy conservation across the instant the string tightens again.The jerk of the string takes away the bob's motion along the string, and energy is lost there.