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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}

We are interested in the Moment of Inertia about a point on the edge (although along a parallel Axis). we can easily determine this using the [Parallel Axis Theorem] :

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

The angular rate of rotation about this point is the same as the angular rate of rotation about the center, {*}ω{*}. The angular momentum is thus

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

The direction of this angular momentum vecor will be pointing downwards at an 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}

This has magnitude

{latex}\begin{large} \[ \tau = r m g\; sin(\theta) \[\end{large}{latex}

 

The top is spinning about its axis with [angular velocity] ω The [moment of inertia] about its axis of rotation is *I* . For the case where the top has its axis perfectly vertical, the angular momentum is given by:

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

The direction associated with both the angular velocity and the angular momentum is directly upwards.

If the axis of the top is tipped from the vertical by an angle θ , then the situation is different. The angular velocity and the angular momentum both point along this angle θ . We define the vector *r* that runs from the point of contact (between the top and the surface it's spinning on) and the [center of mass] of the top. The force of [gravity], *F{~}g{~}* , pulls on the center of mass, and this exerts a [torque (single-axis)] on the top. The torque, τ , is given by the [cross product] between *r* and *F{~}g{~}* :

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

where

{latex}\begin{large}\[ F_{\rm g} = mg \]\end{large}{latex}

and, as the gravitational force, points straight downwards.

|!Spinning Top Precession.PNG!|

The magnitude of the cross product is given by


{latex}\begin{large}\[ \mid \vec{r} \times \vec{F_{\rm g}} \mid = rmg\; sin(\theta) \]\end{large}{latex}

The direction associated with this torque is horizontal and perpendicular to the angular momentum vector *L*. As a result, the torque, which causes a change in the angular momentum vector, does not cause a change in the magnitude of the angular momentum, but only in its direction. The angular momentum vector remains tipped at the angle θ with respect to the vertical, but it begins to rotate around the vertical axis in a counter-clockwise direction with an angular velovitu Ω . This motion is called *precession*.


This new rotational motion ought to contribute to the angular momentum of the system. in most cases of interest and practical importance, however, the small addition to the angular momentum this causes is negligible compared to the angular momentum of the top rotating at angular velocity ω , and so it is usually ignored in the *gyroscopic approximation*, which holds that

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

as long as 

{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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