It is winter Olympics time and time for physics. The event that I always gets me thinking about physics is short track speed skating. It is quite interesting to see these skaters turn and lean at such high angles. All it needs is a little sprinkling of physics for flavor.
Check out this image of Apolo (apparently, it is not Apollo).
How about I start with a force diagram?
I know what you are thinking…Fcent….what force is that? Yes, I am going to use the centrifugal force in this case – but remember that sometimes fake forces are awesome. In short, if I want to pretend like Apolo is not accelerating then I need to add the fake centrifugal force (which is in the opposite direction as the actual acceleration). Remind me later and I will re-visit this problem without using fake forces. Anyway, the centrifugal force will have the magnitude:
Here v is the speed that Apolo (or any skater) is moving and r is the radius of the circle he (or she) is moving in. I drew this vector for the centrifugal force as acting at the center of mass of the skater. This isn’t exactly true. The problem is that different parts of the skater are moving in circles of different radii. However, the difference probably (but I will look at it later) not that large that it matters.
For the other forces, notice that the ice exerts two forces (well, one force that I broke into two components). There is a component parallel to the ice. This is a static friction force where the skate blades cut into the ice. Also, the ice pushes up perpendicular to the ice. This is the normal force. I assumed that the kinetic friction force (which would be into the page opposite the direction of motion) is small enough to be ignored. Really, that is the cool thing about ice skating. Ice needs to be low friction in the direction the skate moves and high friction perpendicular to the blade.
Back to the force diagram, there are two things to consider. The forces must add up to the zero vector (because I am assuming the reference frame of the skater is not accelerating). Also, the torque must be zero about any point. For this case, I will choose the point where the skates touch the ice. This will give three scalar equations (two for the forces an one for the torque). Forgive me, but I am not going to go into all the torque details for now. Wait – I forgot one more parameter – the distance from the point where the skates contact the ice to the center of mass. I will call this distance s (for no particular reason).
Now I can make a substitution for both the centrifugal force (which I wrote above) and the frictional force. I will assume a coefficient of static friction of mus. I will also assume that the skater is just at the point of slipping. This means that the static frictional force is the greatest it can be (so there will be an equal sign and not a less-than-or-equal sign)
Substituting in for the friction and the centrifugal force in the x-direction force equation:
And again for the y-direction substituting for the centrifugal force:
There is another important relationship here. I am going to assume that the sum of the frictional and normal force must be directed towards the center of mass. This means that:
And now, using this in the x-direction force equation to eliminate mu. I get:
This gives me the speed of the skater in terms of the angle he (or she) is at and the radius of the circle the skater is moving in. It turns out I get the same thing if I solve the y-direction force equation (and that would have been a little simpler). Does this result seem reasonable?
- Do the units work? If g is in N/kg (same as m/s2), then g*r will be m2/s2. When I take the square root of this, I get units of m/s – that is good.
- If r is constant, what should happen as theta gets larger? This should be slower speed. It is sort of difficult to see from that function, so let me make a quick plot.
That plot looks pretty good. For an angle that approaches 90 degrees, the skater’s speed would be smaller. A skater wouldn’t have to lean at all if the skater was stopped. As the angle gets smaller (approaching zero), the skater would have to be going faster and faster. That is just what that graph shows.
So, let me see if this works. What is the radius of a short-track? According to the ISU (International Sk8ing Union) the inner radius must be
25-26 meters 8 – 8.5 meters (see correction in comments). What about the angle? From the picture of Apolo above, I get about 33 degrees (0.6 radians). Using these values, I get a speed of:
This is the part where I discuss why the speed was too great. I originally used the 25 meter radius and calculated a speed of around 42 mph. I will leave this part in here even though it is incorrect.
Clearly, this is way too fast. Apolo’s best time on the 500 meter race is 41.5 seconds. This gives an average speed of 12 m/s. Is the angle the problem? I don’t think so. Looking at the plot of the function above, the angle would have to be around 50-60 degrees for the speed to be 12 m/s. Is it because he is pushing on the ice with his hand? Again, I don’t think this is the case because sometimes they don’t touch the ground. What about the radius? Even moving the radius down to 23 meters doesn’t make that big of a difference.
The problem must be with one of my assumptions. I suspect the assumption that the “center of the centrifugal force” was at the same location as the center of mass. This would make a difference. Now I guess I will have to calculate that.
You should give me some credit for knowing something was wrong. According to commenter Milan, the radius is around 8.5 – 8 meters. You can take off some points in my internet looking-up scores. This gives a speed of 24 mph – that I am much happier with. I will fix the figures above.