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1978 Silver Dollar Reverse |
A Coin rolling on its edge with a slight tilt will trace out a circle. What is its radius?
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?
Solution
System:
Interactions:
Model:
Approach:
Diagrammatic Representation
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.
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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).
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Mathematical Representation
The moment of inertia of the coin about its center (on an axis peropendicular to the coin's plane)is
\begin
[ I_
= \frac
m r^
]\end
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 :
\begin
[ I_
= I_
+ m r^
]\end
or
\begin
[ I_
= \frac
m r^
+ m r
= \frac
m r^
]\end
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
\begin
[ L = I \omega = \frac
mr^
\omega ]\end
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:
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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)
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The torque is given by
\begin
[ \vec
= \vec
\times \vec{F_{\rm g}} ]\end
This has magnitude
\begin
[ \mid \vec
\times \vec{F_{\rm g}} \mid = rmg\; sin(\theta) ]\end
This torque is equal to the rate of change of the angular momentum L with time, and the direction is horizontal (by the right-hand rule), which is perpendicular to the direction of the angular momentum. It therefore does not cause a change in the magnitude of the angular momentum, but only in its direction – it causes the angular momentum vector to rotate.
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
\begin
[ L = I \omega ]\end
as long as
\begin
[ \Omega \ll \omega ]\end
In that case,the change in angular momentum only affects the horizontal portion:
and the change is
\begin
[ |\frac{\Delta \vec{L}}
| = L \Omega sin(\theta) = rmg \; sin(\theta) ]\end
and the precession angular velocity is
\begin
[ \Omega = \frac
= \frac
]\end