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CLP-4 Vector Calculus

Section 1.7 Sliding on a Curve

We are going to investigate the motion of a particle of mass \(m\) sliding on a frictionless 1 , smooth curve that lies in a vertical plane. We will consider three scenarios:
  • First, to set things up we'll look at a bead sliding on a stiff wire.
  • Then, we will imagine that we are skiing straight downhill and ask “Where on the hill can we become airborne?”.
  • Then we will imagine that we are skateboarding in a culvert (a large pipe) and ask “When is it safe?”.

Subsection 1.7.1 The Sliding Bead

First, consider a bead of mass \(m\) that is sliding, without friction, on a stiff wire. According to Newton's law of motion
\begin{equation*} m\va = \vF \end{equation*}
where \(\vF\) is the net force being applied to the bead. The bead is subject to two forces. The gravitational force is \(-mg\hj\text{.}\) By definition, absence of friction means that the wire is does not apply any force that is in the direction tangential to the wire. But, because it is stiff, the wire never changes shape and instead applies just the right amount of force, in the direction normal to the wire, that is needed to keep the bead on the wire 2  without bending the wire. Call this normal force \(W\hat\vN\text{.}\)
So, by Newton's law,
\begin{gather*} m\,\va=-mg\,\hj+W\,\hat\vN \end{gather*}
We'll analyse this equation by splitting it into its tangential and normal components.
To extract the tangential component of Newton's law, we dot it with \(\vv=|\vv|\hat\vT\text{.}\) Since \(\hat\vT\cdot\hat\vN=0\) this kills all normal components.
\begin{align*} m\vv\cdot\diff{\vv}{t} &=-mg\hj\cdot\vv+W\hat\vN\cdot\vv\\ \frac{1}{2}m\diff{\ }{t}(\vv\cdot\vv)&=-mg\diff{y}{t} \end{align*}
Here we have used
  • Theorem 1.1.3.c on the left hand side and
  • that \(\hj\cdot\vv\) is just the \(y\) component of \(\vv\) and
  • that \(\hat\vN\) and \(\vv=|\vv|\hat\vT\) are perpendicular.
Moving everything to the left hand side of the equation gives
\begin{align*} \diff{\ }{t}\left(\frac{1}{2}m|\vv|^2+mgy\right)&=0 \end{align*}
and we conclude that the quantity
is a constant, independent of time. This is, of course, the principle of conservation of energy. It determines the speed \(|\vv|=\sqrt{\frac{2E}{m}-2gy}\) of the bead as a function of the height \(y\) (and of the energy \(E\text{,}\) which is determined by the initial conditions).
To extract the normal component of Newton's law, we dot it with \(\hat\vN\text{:}\)
\begin{equation*} m\va\cdot \hat\vN=-mg\hj\cdot \hat\vN+W \end{equation*}
Since
\begin{equation*} \va=\difftwo{s}{t}\,\hat\vT+\ka\Big(\diff{s}{t}\Big)^2\hat\vN =\difftwo{s}{t}\,\hat\vT+\ka|\vv|^2\hat\vN \end{equation*}
and \(\hat\vT\) and \(\hat\vN\) are perpendicular, this gives, after a little rearrangement,

Subsection 1.7.2 The Skier

The difference between the bead on the wire and the skier on the hill is that while the hill is capable of applying an upward normal force (i.e. it can push you upward to keep you from falling to the centre of the Earth), it is not capable of applying a downward normal force. That is the hill cannot pull down on you to keep you on the hill. Only gravity can keep you grounded. There are two main possibilities 3 .
  • If the hill is concave downward as in the figure on the left above, then \(\hat\vN\) points downward and the hill is allowed to have \(W\le 0\) (which corresponds to the normal force \(W\hat\vN\) pushing upward). If ever \(W \gt 0\text{,}\) the hill would have to pull on you to keep you on hill. It can't, so you become airborne. Since \(\hj\cdot \hat\vN \lt 0\text{,}\) this happens whenever
    \begin{align*} W \gt 0 &\iff m\ka|\vv|^2+mg\hj\cdot \hat\vN \gt 0 \iff |\vv| \gt \sqrt{\frac{g}{\ka}|\hj\cdot \hat\vN|} \end{align*}
  • If the hill is concave upward as in the figure on the right above, then \(\hat\vN\) points upward and the hill is allowed to have \(W\ge 0\) (which corresponds to the normal force \(W\hat\vN\) pushing upward). Since \(\hj\cdot \hat\vN \gt 0\) we always have \(W=m\ka|\vv|^2+mg\hj\cdot \hat\vN \gt 0\text{.}\) You never become airborne. On the other hand your knees may complain.

Subsection 1.7.3 The Skate Boarder

So far, Equations 1.7.1 and 1.7.2 apply to any stiff frictionless “wire”. We now specialize to the special case of a skateboarder inside a circular culvert of radius \(a\text{.}\) Let's put the bottom of the circle at the origin \((0,0)\text{,}\) so that the centre of the circle is at \((0,a)\text{.}\)
In this case the curvature is \(\ \ka=\frac{1}{a}\ \) and \(\hj\cdot\hat\vN=\cos \phi=\frac{a-y}{a}\) so 1.7.1 and 1.7.2 simplify to
\begin{align*} |\vv|&=\sqrt{\frac{2}{m}(E-mgy)}=\sqrt{2g\Big(\frac{E}{mg}-y\Big)}\\ W&=\frac{2}{a}(E-mgy)+\frac{mg}{a}(a-y) =\frac{3mg}{a}\Big(\frac{2}{3mg}E+\frac{a}{3}-y\Big) \end{align*}
Imagine now that you start at the bottom of the culvert, that is at \(y=0\text{,}\) with energy \(E \gt 0\text{.}\) As time progresses, \(y\) increases and consequently \(|\vv|\) and \(W\) both decrease, as, of course, they should. This continues until one of the following three things happen.
  1. \(|\vv|\) hits 0, in which case you stop rising and start descending. The speed \(|\vv|\) is zero when \(y=y_S=\frac{E}{mg}\text{.}\) (The subscript “\(S\)” stands for “stop”.) Physicists say that when you reach \(y_S\) all of your kinetic energy (\(\frac{1}{2}m|\vv|^2\)) has been converted into potential energy (\(mgy\)).
  2. \(W\) hits zero. When you get higher than this, \(W\) becomes negative and the culvert would have to pull on you to keep your feet on the culvert. As the culvert can only push on you, you become airborne. The normal force \(W\) is zero when \(y=y_A=\frac{2}{3}\frac{E}{mg}+\frac{a}{3}\text{.}\) (The subscript “\(A\)” stands for “airborne”.)
  3. \(y\) hits \(2a\text{.}\) This is the summit of the culvert. You descend on the other side.
Which case actually happens is determined by the relative sizes of \(y_S,\ y_A\) and \(2a\text{.}\)
  • Comparing \(y_S=\frac{2}{3}\frac{E}{mg}+\frac{1}{3}\frac{E}{mg}\) and \(y_A=\frac{2}{3}\frac{E}{mg}+\frac{a}{3}\text{,}\) we see that \(y_S\le y_A\iff \frac{E}{mg}\le a \text{.}\)
  • Comparing \(y_A=\frac{2}{3}\frac{E}{mg}+\frac{a}{3}\) and \(a=\frac{2}{3}a+\frac{a}{3}\text{,}\) we see that \(y_A\le a\iff \frac{E}{mg}\le a\text{.}\)
  • Comparing \(y_A=\frac{2}{3}\frac{E}{mg}+\frac{a}{3}\) and \(2a=\frac{5}{3}a+\frac{a}{3}\text{,}\) we see that \(y_A\le 2a\iff \frac{E}{mg}\le \frac{5}{2}a\text{.}\)
So the conclusions are:
  • If \(\ {\bf 0\le\frac{E}{mg}\le a}\ \) then \(\ 0\le y_S\le y_A\le a\ \text{.}\) In this case you just oscillate between heights 0 and \(y_S\le a\) in the bottom half of the culvert, as in the figure on the left below.
  • If \(\ {\bf a\le\frac{E}{mg}\le \frac{5}{2}a}\ \) then \(\ a\le y_A\le y_S,2a\ \text{.}\) In this case you make it more than half way to the top. But you become airborne at \(y=y_A\) which is somewhere between the half way mark \(y=a\) and the top \(y=2a\text{.}\) At this point our model breaks down because you are no longer in contact with the culvert. You just freely follow a parabolic arc until you crash back into the culvert, as in the figure in the centre below.
  • If \(\ {\bf \frac{5}{2}a \lt \frac{E}{mg}}\ \) then \(\ 2a \lt y_A \lt y_S\ \text{.}\) In this case you successfully go all the way around the culvert, looping the loop, as in the figure on the right below. Note that, as \(\frac{E}{mg} \gt \frac{5}{2}a \gt 2a\text{,}\) this requires significantly more energy than that required to reach the top, i.e. to reach height \(2a\text{.}\)

Exercises 1.7.4 Exercises

Exercise Group.

Exercises — Stage 1
You may assume the acceleration due to gravity is \(g=9.8\) m/s\(^2\text{.}\) You may also assume that the systems described function as they do in the book: so tracks are frictionless, etc., unless otherwise mentioned.
1.
The figure below represents a bead sliding down a wire. Sketch vectors representing the normal force the wire exerts on the bead, and the force of gravity.
Assume the top of the page is “straight up.”
2.
In the definition \(E=\frac12m|\vv|^2+mgy\text{,}\) \(\vv\) is the derivative of position with respect to what quantity?
3.
A bead slides down a wire with the shape shown below, \(x \lt 0\text{.}\)
Let \(W\hN\) be the normal force exerted by the wire when the bead is at position \(x\text{.}\) Note \(W \gt 0\text{.}\) Is \(\diff{W}{x}\) positive or negative?
4.
A skateboarder is rolling on a frictionless, very tall parabolic ramp with cross-section described by \(y=x^2\text{.}\) Given a boarder of mass \(m\) with system energy \(E\text{,}\) what is the highest elevation the skater reaches? How does this compare to a circular culvert?

Exercise Group.

Exercises — Stage 2
5.
A skateboarder of mass 100 kg is freely rolling in a frictionless circular culvert of radius 5 m. If the skateboarder oscillates between vertical heights of 0 and 3 m, what is the energy \(E\) of the system?
6.
A skateboarder is rolling on a frictionless circular culvert of radius 5 m. What should their speed be when they're at the bottom of the culvert (\(y=0\)) for them to make it all the way around?
7.
A ball of mass 1 kg rolls down a track with the shape \(\vr(\theta)=(3 \cos \theta, 5\sin\theta, 4+4\cos\theta)\) for \(0 \le \theta \le \frac{\pi}{2}\text{.}\) Coordinates are measured in metres, and the \(z\) axis is vertical (so the force due to gravity is \(-mg\hk\text{.}\))
When \(\theta=\pi/4\text{,}\) the particle has instantaneous velocity \(|\vv(t)|=5\) m/s. What is the normal force exerted by the track at that time? Give your answer as a vector.
8.
A bead of mass \(\frac{1}{9.8}\) kg slides down a wire in the shape of the curve \(\vr(\theta)=(\sin \theta , \sin \theta - \theta)\text{,}\) \(\theta \ge 0\text{,}\) with coordinates measured in metres. The bead will break off the wire when the wire exerts a force of 100 N on the bead.
If the bead breaks off the wire at \(\theta=\frac{13\pi}{3}\text{,}\) how fast is the bead moving at that point?
9.
A skier is gliding down a hill. The hill can be described as \(\vr(t)=(\ln t, 1-t)\text{,}\) \(1/e \le t \le e\text{,}\) with coordinates measured in kilometres. How fast would the skier have to be moving in order to catch air?

Exercise Group.

Exercises — Stage 3
10.
A wire follows the arclength-parametrized path \(\vr(s)=(x(s),y(s))\text{.}\) A bead, equipped with a jet pack, slides down the wire. The jet pack can exert a variable force in a direction tangent to the wire, \(U\hT\text{.}\) Assuming the bead slides with constant speed \(\left|\diff{\vr}{t}\right|=c\left| \diff{\vr}{s}\right|=c\text{,}\) find a simplified equation for \(U\text{,}\) the signed magnitude of the force exerted by the jet pack.
Let the acceleration due to gravity be \(g\text{,}\) and let the mass of the bead with its jet pack be \(m\text{.}\) Give \(U\) as a function of \(s\text{.}\)
Remark: most beads this author has seen did not have jet packs. However, in modelling a frictionful 4  system, friction acts as a force that is directly opposing the direction of motion — much like our jet pack.
11.
A snowmachine is cautiously descending a hill in low gear. Its engine provides a force \(M\hT\) parallel to the direction of motion. The engine provides whatever force is necessary to keep the snowmachine moving at a constant speed, \(|\vv|\text{.}\) Its treads do not slip.
  1. Give a formula for \(M\) in terms of the mass \(m\) of the snowmachine, the acceleration due to gravity \(g\text{,}\) and the tangent vector \(\hT\) to the hill.
  2. Let \(\hT\) point in the downhill direction. Do you expect \(M\) to be positive or negative as the snowmachine moves downhill?
  3. Find \(M\) for the hill of shape \(y=1+\cos x\) (measured in metres) at the point \(x=\frac{3\pi}{4}\) for a snowmachine of mass 200 kg.
12.
A skateboarder rolls along a culvert with elliptical cross-section described by
\begin{equation*} \vr(\theta)=(4\cos\theta,3(1+\sin\theta)), 0 \le \theta \le 2\pi, \end{equation*}
with coordinates measured in metres.
  1. Give the height \(y_S\) (in terms of \(m\text{,}\) \(g\text{,}\) and \(E\)) where the skater's speed is zero.
  2. Write an equation relating \(E\text{,}\) \(m\text{,}\) \(g\text{,}\) and \(y_A\text{,}\) where \(y_A\) is the \(y\)-value where the skater would become airborne, i.e. where \(W=0\text{.}\) (You do not have to solve for \(y_A\) explicitly.)
  3. Suppose the skater has speed 11 m/s at the bottom of the culvert. Which of the following describes their journey: they make it all the way around; they roll back and forth in the bottom half; or they make it onto the ceiling, then fall off?
13.
A frictionless roller-coaster track has the form of one turn of the circular helix with parametrization \(\ (a\cos\theta,a\sin\theta,b\theta).\) A car leaves the point where \(\ \theta=2\pi\ \) with zero velocity and moves under gravity to the point where \(\ \theta=0\text{.}\) By Newton's law of motion, the position \(\vr(t)\) of the car at time \(t\) obeys
\begin{equation*} m \vr''(t) = \vN\big(\vr(t)\big) - mg\hk \end{equation*}
Here \(m\) is the mass of the car, \(g\) is a constant, \(-mg\hk\) is the force due to gravity and \(\vN\big(\vr(t)\big)\) is the force that the roller-coaster track applies to the car to keep the car on the track. Since the track is frictionless, \(\vN\big(\vr(t)\big)\) is always perpendicular to \(\vv(t)=\diff{\vr}{t}(t)\text{.}\)
  1. Prove that \(E(t)=\half m |\vv(t)|^2 +mg \vr(t)\cdot\hk\) is a constant, independent of \(t\text{.}\) (This is called “conservation of energy”.)
  2. Prove that the speed \(|\vv|\) at the point \(\theta\) obeys \(\ |\vv|^2=2gb(2\pi-\theta).\)
  3. Find the time it takes to reach \(\theta=0\text{.}\)
We are mathematicians — we like idealized situations.
This force is required to keep the bead from either passing through the wire or flying off the wire.
We assume that you are going downhill and that the curvature \(\ka\gt 0\text{.}\)
Frictionated? Frictiony? Befrictioned?