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|!1978_dollar_rev.jpg!|
|1978 Silver Dollar Reverse
from Wikimedia Commons|



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{excerpt}A Coin rolling on its edge with a slight tilt will trace out a circle. What is its radius?{excerpt}

A coin is rolling without slipping on its edge, but tilts slightly to one side making an angle *θ* with the vertical, which makes the rolling path trace out a circle. What is the radius of this circle?

 



h4. Solution

{toggle-cloak:id=sys} *System:*  {cloak:id=sys}The Rolling Coin .{cloak}

{toggle-cloak:id=int} *Interactions:*  {cloak:id=int}[torque (single-axis)] due to [gravity] and normal force from the surface the top is spinning on.{cloak}

{toggle-cloak:id=mod} *Model:* {cloak:id=mod} [Angular Momentum and External Torque].{cloak}

{toggle-cloak:id=app} *Approach:*  

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{toggle-cloak:id=diag} {color:red} *Diagrammatic Representation* {color}

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A coin rolling without slipping on a horizontal surface and tipped slightly at an angle θ to the vertical will roll in a circle of radius *R*. What determines this radius, and how does it vary with other characteristic values?

Call the radius of the coin *r* and its mass *m*. [Gravity] pulls on the center of mass with gravitational force *mg* , and the surface the coin is rolling on pushes upwards with a normal force *F*. This must be equal to *mg* , since the coin neither rises nor falls above or below the surface. 

|!Rolling Coin 1.PNG!|

We must choose a point about which to calculate the angular momentum and torque. A convenient spot is the point at which the coin touches the surface. Doing so allows us to neglect the normal force in our calculation of torque. We can also simply consider the rotation of the coin about that point in calculating the angular momentum (If we chose any other point, we'd have to consider the motion of the center of mass of the coin as well).

|!Rolling Coin 2.PNG!|



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{toggle-cloak:id=math} {color:red} *Mathematical Representation* {color}

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The [Moment of Inertia] of the coin about its center (on an axis peropendicular to the coin's plane)is 

{latex}\begin{large} \[ I_{\rm center} = \frac{1}{2} m r^{2} \]\end{large}{latex}

{latex}

or

{latex}\begin{large} \[ I_{\rm edge} = \frac{1}{2}m r^{2} + m r{2} = \frac{3}{2} m r^{2} \]\end{large}{latex}

 {*}ω{*}. The angular momentum is thus

{latex}\begin{large} \[ L = I \omega = \frac{3}{2}mr^{2} \omega \]\end{large}{latex}

 angle {*}θ{*} with respect to the vertical, since the coin is tipped by that angle from the vertical:


|!Picture of coin in profile showing angular momentum!|

The [torque] is due to the force of gravity acting on the center of mass of the coin (since there is no torque about the contact point due to the normal force, as the distance is zero)

|!Picture of coin in profile showing torque!|

The torque is given by

{latex}\begin{large} \[ \vec{\tau} = \vec{r} \times \vec{F_{\rm g}} \]\end{large}{latex}

\begin{large} \[ | \tau | =  \left| \frac{dL}{dt} \right| = \Omega L\; cos(\theta) = rmg\; sin(\theta) \]\end{large}

\begin{large} \[ m g r\; sin(\theta) = \frac{3}{2} m r v\; \Omega \; cos(\theta) \]\end{large}

\begin{large}\[ v = \omega r = \Omega R \]\end{large}

\begin{large} \[ R = \frac{3 v^{2}}{2g\; tan(\theta)}\]\end{large}

\begin{large}\[ L = I \omega \]\end{large}{latex}

 

{latex}\begin{large}\[ \Omega \ll \omega \]\end{large}{latex}

In that case,the change in angular momentum only affects the horizontal portion:


and the change is


{latex}

\begin{large} \[ |\frac{\Delta \vec{L}}{\Delta t} | = L \Omega sin(\theta) = rmg \; sin(\theta) \]\end{large}{latex} 

and the precession angular velocity is 

{latex}\begin{large} \[ \Omega = \frac{rmg}{L} = \frac{rmg}{I\omega}  \]\end{large}{latex}

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r \ll R \]\end{large}