# Why do rolling disc (coin) move in circular path?

We have a coin that is rolled such that it's tilted at an small angle $\theta$.

Question:: What turns around rolling disc so that it traces circular motion (spiral as it's speed decreses)?

Problem statement:: In the following figure, what turns the disc so that it follows the circular path? Solution::
I interpreted the problem incorrectly. Only a sliding body would have a torque like that.

• What is $J$ here? Is torque in diagram spinning the coin? Aug 6, 2012 at 10:36
• @Ankush since the coin is moving in circle ... something must continuously change the direction of the coin (so that coin follows circular path).
– S L
Aug 6, 2012 at 11:01
• That doesn't answer my comment. Is $J$ force you expect? If yes, why outside? The torque is misleading too. It's direction means coin is spinning. Like earth rotating. But you want answer for revolution. Right? Aug 6, 2012 at 11:18
• $J$ is angular momentum.
– S L
Aug 6, 2012 at 11:27
• @Ankush The coin is moving in circle means it should be spinning ... with period equal to the period of circular motion.
– S L
Aug 6, 2012 at 11:28

I'm not sure if this is the dominant factor, but...

Once the coin begins to tip at all, there is torque due to gravity. If you work it out with your hands you'll see that this torque acts perpendicular to the angular momentum from the rolling of the coin and in the plane it rolls on. Thus, it acts to direct the coin in a circular path.

A quick experiment shows the coin will travel in a straight line unless it is tipped, so I think this is what's going on.

Nice pictures by the way!

• can you tell me if the Torque I showed in the figure is right?
– S L
Aug 4, 2012 at 19:53
• The torque is zero when the coin is straight up and down. Aug 4, 2012 at 19:54
• sorry ... i couldn't find the tool to rotate the disc earlier
– S L
Aug 4, 2012 at 19:55
– S L
Aug 4, 2012 at 20:05
• The torque is not downward. It's in the plane the coin rolls on and perpendicular to its angular momentum. This is why it makes a circle. Aug 4, 2012 at 20:17

Consider a bucket which is quite stable having and has $r_{1} < r_{2}$. Note circumference $2\pi r_{1}$ is less than outer circumference $2\pi r_{2}$.Now if bucket is set into motion, the point which lies on inner circumference has to cover less distance compared to point on outer circumference. Since whole body is rigid and stable, reaction forces because of weight of object change path of object into running into circle. Note when running in circle, inner point lies close to this circle's center and thus covering less distance.

For thinner disks like coins, the effect is because of tilt. Here torque and frictional forces guide the path.

Update: Question has changed for coin case only. Consider this free body diagram. Here $\vec{N}$ is normal force and $\vec{G}$ is gravitational force. These two create a torque $\vec{\tau}$ about any point in the frame of reference. This $\vec{\tau}$ is perpendicular to the picture going inside (clockwise). Since coin is rolling on plane, the ground's frictional $\vec{f_{1}}$force prevents it from toppling. Had the surface been super smooth the coing will just fall and won't run in circle. Hence it's the frictional force which causes circular path.

Note $\vec{f_{1}}$ exerts equal reverse torque to prevent coin from falling. If it fails, coin will fall.

– S L
Aug 5, 2012 at 22:38
• @experimentX updated the answer. Aug 6, 2012 at 6:10
• shouldn't this kind of torque make the coin fall? ... it comes towards us.
– S L
Aug 6, 2012 at 9:37
• @experimentX Fall: No. Because $f_{1}$ exerts reverse torque. Aug 6, 2012 at 10:04
• wouldn't that torque be in out direction? please check update.
– S L
Aug 6, 2012 at 10:19

Think of a larger object in this situation, like a motorcycle, the centrifugal force wants to keep it upright and the force of gravity want's to pull it down (in the direction it is leaning, and the centrifugal force is converted to friction in this case.) Take a look at this video.

Funnel Physics

Bicycle and motorcycle dynamics

Spinning coin inside a ballon- centripetal and centrifugal forces

Your question asks why there is a rotatory force that turns the coin to the right. The gravitational force is perpendicular to surface and so the vector cross-product of the gravitation force and the angular momentum (which is not exactly parallel to the surface) is perpendicular to both those vectors. That doesn't actually help (for me anyway). However, if you think of the leading edge and trailing edge of the coin as distinct entities, their velocities are opposite, so the leading edge experiences a resultant force (cross-product of V and g with a constraint imposed by the surface) inward and the trailing edge experiences an opposite force outward and so the coin turns to the right.

$$\vec{f_{1}}$$ does not create a torque to cancel the weight's torque. Torque is $$F\times r$$, all torques have to be calculated w. respect to same point. If weight's torque was calculated w. respect to the ground, $$\vec{f_{1}}$$'s should be too. In that case, $$r=0$$ and $$f_1$$'s torque is zero.

In fact, the weight's torque is unbalanced. It is directed into the page. If the coin isn't rolling, this torque makes it fall.

But if it is rolling, it has an angular momentum $$\vec{L}$$, and $$\vec{\tau}=\Delta \vec{L}$$. If the coin is rolling toward us (out of the page) then $$\vec{L}$$ is to the right (and a bit tilted, but the right-hand component is greater and is the one that matters here). Since $$\Delta \vec{L}$$ is in the direction of the torque (into the page), then the effect of the unbalanced torque is to shift the direction of $$\vec{L}$$ a little bit into the page. That means the angular momentum rotates.