﻿ Solution: Particle on Frictionless Track | Primer on Geometric Algebra | David Hestenes

## Solution: Particle on Frictionless Track

A particle slides on the frictionless track below subject to a force  mdv / dt  that keeps it on the track. Sketch its velocity and acceleration at the indicated points.

The
velocities are sketched below with blue arrows. Their directions are trivial, since they are always tangent to the path. Intuitively, you might correctly guess that the particle speeds up going downhill and slows down going uphill, so the blue vectors should be longer at lower elevations.

We can pin down the relationship between speed and elevation by working with the starting equation. Since
N is always perpendicular to v, dotting both sides by v simplifies and leads us to the familiar conservation of energy theorem. Certainly g and v0 offer two natural constant directions. We may as well define the elevation at point to be y 0 and the elevation at the start (or point or point ) to be y0, which is also 2R (the diameter of the hoop).

$\Large{\begin{array}{l}\left( {m{\bf{g}} + {\bf{N}}\,\,\, = \,\,\,m\frac{{d{\bf{v}}}}{{dt}}} \right) \cdot {\bf{v}}\,\,\,\, \Rightarrow \,\,\,\,m{\bf{g}} \cdot \frac{{d{\bf{x}}}}{{dt}} + 0\,\,\, = \,\,\,\frac{1}{2}m\left( {{\bf{v}}\frac{{d{\bf{v}}}}{{dt}} + \frac{{d{\bf{v}}}}{{dt}}{\bf{v}}} \right)\\ ~~~~~~~~\Rightarrow \,\,\,\,\frac{d}{{dt}}\left( {\frac{1}{2}m{v^2} - m{\bf{g}} \cdot {\bf{x}}} \right)\,\,\, = \,\,\,0\,\,\,\, \Rightarrow \,\,\,\,\frac{1}{2}m{v^2} - m{\bf{g}} \cdot {\bf{x}}\,\,\, = \,\,\,{\rm{constant.}}\\\\\\{\rm{KE}}\,\,\, \equiv \,\,\,\frac{1}{2}m{v^2},\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{PE}}\,\,\, \equiv \,\,\, - m{\bf{g}} \cdot {\bf{x}}\,\,\, = \,\,\,mgy.\\\\{\left( {\frac{1}{2}m{v_0}^2 + mg{y_0}} \right)_{{\rm{top}}}}\,\,\, = \,\,\,\,{\left( {\frac{1}{2}m{v^2} + mgy} \right)_{{\rm{anywhere}}}}\end{array}}$

Therefore, we can see how speed varies with elevation:

$\Large{\begin{array}{l}v\,\,\, = \,\,\,\sqrt {{v_0}^2 + 2g\left( {{y_0} - y} \right)} \,\,\, = \,\,\,\sqrt {{v_0}^2 + 4gR\left( {1 - y/{y_0}} \right)} \,\,\,\\\\\\{v_0} = {v_6} = {v_9}\,,\,\,\,\,\,~~~~\,\,{v_1} \approx \sqrt {{v_0}^2 + 0.4gR} \,,\,\,\,\,~~~~\,\,\,{v_4} = \sqrt {{v_0}^2 + 4gR} \,,\\\\{v_2} \approx {v_8} \approx {v_5} = {v_7} = \sqrt {{v_0}^2 + 2gR} \,,\,\,\,\,~~~~\,\,\,{v_3} \approx \sqrt {{v_0}^2 + 3gR} \end{array}}$

The accelerations are sketched with red arrows. Of course they are not (can't be; units are different) to the same scale as the blue velocities, so we are just comparing the sizes and directions of the red arrows relative to each other. The direction of an acceleration can be estimated by seeing how the blue velocity vector is changing, that is by considering ∆v, because  a = dv / dt.  For point , for example, notice that v is rotating CW and growing, since the particle is sliding downhill:

This logic also explains why the acceleration vector points inside a curving path. The normal component of acceleration is often called the centripetal or radial component, a. Of course, there will also be a component along the path, a|| , if the speed is changing. We can combine the general expression for acceleration given earlier with the equation of motion for this example to get:

$\Large{\begin{array}{l}{\bf{a}}\,\,\, = \,\,\,\frac{{d{\bf{v}}}}{{dt}}\,\,\, = \,\,\,{\bf{\hat v}}\frac{{dv}}{{dt}}\,\, + \,\,{{{\bf{\hat a}}}_R}\,\,\frac{{{v^2}}}{r}\,\,\, = \,\,\,{\bf{g}}\,\, + \,\,\frac{{\bf{N}}}{m}\,\,\, = \,\,\,{{\bf{g}}_\parallel }\,\, + \,\,\left( {{{\bf{g}}_ \bot }\,\, + \,\,\frac{{\bf{N}}}{m}} \right)\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left| {\frac{{dv}}{{dt}}} \right|\,\,\, = \,\,\,\left| {{{\bf{g}}_\parallel }} \right|\,,\,\,\,\,\,\,\,\,~~~~~~~~\,\,\,\,\,\,\,\,\,\,\,\,\frac{{{v^2}}}{r}\,\,\, = \,\,\,\left| {{{\bf{g}}_ \bot } + \frac{{\bf{N}}}{m}} \right|\end{array}}$

Notice that g and N are in the same direction at point and in opposite directions at point . The radial acceleration (v2/r) is affected greatly in our example by the radius of curvature, r, of our frictionless path. We saw above that the speed increases as the elevation decreases, but r varies more wildly because of the shape of our path. In the circular loop, r = R. At points , , , and , r = ∞, so the radial acceleration is zero. And, at point , we can estimate rR/2. With these ideas, you may now agree with the qualitative features of the red arrows in the diagram above.

There's no need for further detail here, except to compare the two points  and  at the bottom and top of a vertical frictionless circle. These points are often compared in textbooks for a track as above or a cyclist in an amusement park exhibit or a rock attached to a string. What initial speed does the cyclist need so that the cycle maintains contact with the track at the top (N > 0)? How fast can we twirl the rock so that maximum tension (N at the bottom) doesn't break the string?

$\Large{\begin{array}{l}{{\bf{a}}_{(4)}}\,\, = \,\,\,{{\bf{a}}_{R(4)}}\,\, = \,\,\, - {\bf{\hat g}}\,\,\frac{{{v_{(4)}}^2}}{r}\,\,\,\, = \,\,\, - {\bf{\hat g}}\,\,\left( {\frac{{{v_0}^2}}{r}\,\, + \,\,4g} \right)\\{{\bf{a}}_{(6)}}\,\, = \,\,\,{{\bf{a}}_{R(6)}}\,\, = \,\,\, + {\bf{\hat g}}\,\,\frac{{{v_{(6)}}^2}}{r}\,\,\,\, = \,\,\, + {\bf{\hat g}}\,\,\frac{{{v_0}^2}}{r}\,\\\\\\{{\bf{N}}_{(4)}}\,\, = \,\,\,m{{\bf{a}}_{(4)}}\,\, - \,\,m{\bf{g}} = \,\,\, - m{\bf{g}}\,\,\left( {\frac{{{v_0}^2}}{{rg}}\,\, + \,\,5} \right)\\{{\bf{N}}_{(6)}}\,\, = \,\,\,m{{\bf{a}}_{(6)}}\,\, - \,\,m{\bf{g}} = \,\,\, + m{\bf{g}}\,\,\left( {\frac{{{v_0}^2}}{{rg}}\,\, - \,\,1} \right)\\\\\\{a_{R(4)}}\,\, = \,\,\,{a_{R(6)}}\,\, + \,\,4g\,,\,\,\,\,\,\,\,\,~~~~~~\,\,\,\,\,\,\,\,\,\,\,\,{N_{(4)}}\,\,\, = \,\,\,{N_{(6)}}\,\, + \,\,6mg\end{array}}$

For the rock on a string being swung at the slowest speed to maintain a vertical circle, the string had better be at least strong enough to hold six such rocks! We can see the circular track had better also use some sturdy construction at the bottom. At first, it may seem strange that the centripetal acceleration at the bottom is only 4
g more than at the top while the normal force at the bottom is 6mg more than at the top, but that's because, as mentioned above, g and N point in the same direction at the top and in opposite directions at the bottom.