Turn a round steel bowl upside down and set a small ice cube at the top. Give it the gentlest touch. It slides down the curve, faster and faster. It does not slide all the way down the side. Partway down, it leaves the steel and flies off through the air.
Sliding down the outside
Take a small body of mass \(m\) on the outside of a smooth dome of radius \(R\), with centre O. On a dome the natural angle is measured from the top. So in this lesson \(\varphi\) is the angle at O between the upward vertical and the radius to the body. At the top \(\varphi = 0\). The body has then come down by \(R(1 - \cos\varphi)\).
The angle \(\varphi\) of this lesson is measured from the top, not from the bottom like \(\theta\) in Lessons 1 to 3. Mixing the two changes \(\cos\) into \(-\cos\) and ruins every result. Check any formula at \(\varphi = 0\), where the body is at the top.
Two forces act on the body: its weight \(mg\) and the normal force \(N\) of the dome, which points outwards along the radius. The weight splits into \(mg\cos\varphi\) along the radius, towards O, and \(mg\sin\varphi\) along the surface, down the slope. The part along the surface speeds the body up. The radial parts must supply \(mv^2/R\), towards O.
\[ mg\cos\varphi - N = \dfrac{mv^2}{R} \]
Why. towards O, the weight's radial part minus the outward normal force must equal the needed inward force.
\[ v^2 = 2gR(1 - \cos\varphi) \]
Why. the dome is smooth and \(N\) does no work, so energy is conserved. Starting from rest at the top, the body has fallen \(R(1 - \cos\varphi)\).
\[ N = mg\cos\varphi - 2mg(1 - \cos\varphi) \]
Why. put \(v^2\) into step 1 and solve for \(N\).
\[ N = mg(3\cos\varphi - 2) \]
Read it. collect the terms in \(\cos\varphi\).
The normal force falls for two reasons. As the body moves down, \(\cos\varphi\) falls, so less of the weight presses into the dome. At the same time the body speeds up, so the circle needs more of the weight's radial part to turn it. Both take something away from \(N\).
A \(2\,\text{kg}\) block slides from rest at the top of a smooth dome. Take \(g = 10\,\text{m/s}^2\) and \(\cos 37° = 0.8\). What is the normal force, in newtons, when \(\varphi = 37°\)?
Where it leaves
A dome can push but cannot pull, so \(N \ge 0\). The body stays on the surface while \(3\cos\varphi - 2 \ge 0\). The normal force becomes zero where \(\cos\varphi = 2/3\), that is \(\varphi \approx 48.2°\). Beyond that point the dome would have to pull the body inwards. It cannot, so the body leaves the surface there and becomes a projectile.
A small block is given a gentle nudge at the top of a smooth dome. Where does it leave the surface?
The height of the leaving point above the centre is \(R\cos\varphi = 2R/3\), a third of the radius below the top. The speed there is \(v^2 = 2gR(1 - 2/3) = 2gR/3\). The angle does not depend on \(m\), because every force in the radial equation is proportional to \(m\). It does not depend on \(g\) either. On the Moon the block would slide more slowly, but it would leave at the same point.
In the ice-cube test, a bowl of radius \(10\,\text{cm}\) lets the cube go about \(3.3\,\text{cm}\) below the top, at \(48°\). From there the cube flies clear of the steel. Any real bowl has a little friction, so the cube leaves a little lower, but the smooth result is close.
A smooth hemisphere of radius \(3\,\text{m}\) rests with its flat face on the ground. A small block slides down from its top after a gentle nudge. At what height, in metres above the ground, does it leave the surface?
Two blocks, one of \(1\,\text{kg}\) and one of \(5\,\text{kg}\), are nudged from the top of the same smooth dome. Which leaves the surface first, higher up?
A small block slides from rest down a smooth sphere of radius \(1.5\,\text{m}\). Take \(g = 10\,\text{m/s}^2\). What is its speed, in m/s, at the instant it leaves the surface?
A small block is nudged from the top of a smooth hemisphere of radius \(1.5\,\text{m}\) that rests with its flat face on the floor. Take \(g = 10\,\text{m/s}^2\). Where does the block land on the floor?
It leaves where \(\cos\varphi = 2/3\), so \(\sin\varphi = \sqrt5/3 \approx 0.745\). That point is \(1.5 \times 2/3 = 1\,\text{m}\) above the floor and \(1.5 \times 0.745 \approx 1.118\,\text{m}\) to the side of the centre.
\(v^2 = 2gR/3 = 10\), so \(v \approx 3.162\,\text{m/s}\), along the tangent at \(\varphi\) below the horizontal. Its parts are \(v\cos\varphi \approx 2.108\,\text{m/s}\) sideways and \(v\sin\varphi \approx 2.357\,\text{m/s}\) down.
Falling \(1\,\text{m}\): \(1 = 2.357\,t + 5t^2\), so \(t \approx 0.270\,\text{s}\). Sideways it moves \(2.108 \times 0.270 \approx 0.569\,\text{m}\) more.
It lands \(1.118 + 0.569 \approx 1.69\,\text{m}\) from the centre, only \(0.19\,\text{m}\) beyond the edge of the hemisphere.
With a starting speed
Now start the body at the top with a speed \(v_0\). Energy gives \(v^2 = v_0^2 + 2gR(1 - \cos\varphi)\), and the radial equation gives \(N = mg(3\cos\varphi - 2) - mv_0^2/R\). The extra term lowers the normal force everywhere. So the body leaves sooner, higher up the dome, where \(\cos\varphi = 2/3 + v_0^2/(3gR)\).
When \(v_0^2 = gR\), the formula gives \(\cos\varphi = 1\): the body leaves at the very top. At that speed the weight alone turns the body on the dome's own curve, and the dome carries no load at all. This is the car over a crest from the chapter on Dynamics of Circular Motion. A car crossing a hump of radius \(R\) at \(\sqrt{gR}\) or faster leaves the road at the top.
A small block is placed at the top of a smooth sphere of radius \(0.4\,\text{m}\). Take \(g = 10\,\text{m/s}^2\). What is the least horizontal speed, in m/s, that makes it leave the surface at once?
A block starts at \(2.5\,\text{m/s}\) from the top of a smooth dome of radius \(2.5\,\text{m}\). Take \(g = 10\,\text{m/s}^2\). At what height, in metres above the centre, does it leave the surface?
- Measuring the angle from the bottom and using \(\cos\varphi = 2/3\).The result \(\cos\varphi = 2/3\) has \(\varphi\) measured from the top. From the bottom of a full sphere the leaving angle is about \(132°\).
- Setting the speed to zero to find where the body leaves.The body leaves while moving, where the normal force \(N = mg(3\cos\varphi - 2)\) becomes zero.
- Thinking a heavier body leaves sooner.The mass cancels. Every body leaves a smooth dome at \(\cos\varphi = 2/3\), a height \(2R/3\) above the centre.
- Forgetting the starting speed.A body started at \(v_0\) leaves where \(\cos\varphi = 2/3 + v_0^2/(3gR)\), and at the top itself when \(v_0 \ge \sqrt{gR}\).