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Physics on ice

Physics World April 2020

Physics World

 
Lateral Thoughts Physics World  April 2020

Physics on ice

(iStock / timmy)

When I watch films, TV shows or sports I often find myself thinking about the physics of the situation. Here in North America, we’re approaching the end of ice hockey season. The main aim of this sport is simply to get the puck into the net. But how far could a player actually hit a puck, if the net and edge of the rink weren’t there? Could you even make the puck loop all around the rink? These kinds of questions are ideal tools for teaching physics, as you can start with the most basic scenario and build upon it to reach the complex reality.

How far could a player actually hit a puck, if the net and edge of the rink weren’t there? Could you even make the puck loop all around the rink? Such questions are ideal tools for teaching physics

To answer how far you can hit a puck, there are three basic layers. You start with the ice, which is a very slippery surface – so it’s safe to assume that the friction between the puck and the ice is negligible. You also ignore air resistance, which only leaves the downward gravitational force and the normal force (the upward pushing force from the ice), which balance each other out. No net force means no movement, so you apply a pushing force, such as a hit from a hockey stick, which results in the puck travelling at a constant speed forever.

All of this is simple mechanics, but it’s not quite realistic. Although ice is very slippery, there will be a frictional force between it and the puck, which acts against the forward motion – meaning you must account for it. At a fundamental level, friction is a complicated interaction, but these complexities can be captured with a simple model of friction, which is found experimentally for different materials. Assuming the coefficient of friction is about 0.1 for our puck on ice, using some basic kinematics and Newton’s handy laws, that gives a stopping distance of just over 1000 m when the puck is hit with a starting speed of 160 km/hr.

While more realistic than never stopping, this scenario is still not believable, as air resistance also needs to be taken into account. Although the collisions between air molecules and objects can be complex, like with friction there is a model to describe the scenario. Unfortunately, when putting this into the equation for acceleration there’s a snag – the acceleration can be used to determine the change in velocity, but the magnitude of the acceleration now also depends on the velocity.

It’s possible to write the acceleration as the derivative of velocity with respect to time, turning this equation into a differential equation. But there is another route in the form of numerical calculations, which allow you to take a problem and break it down into many smaller and simpler problems. In this case, the motion of a sliding hockey puck can be modelled in small time steps, let’s say 0.1 seconds. During that tenth of a second, the hockey puck will indeed decrease in speed. However, the change in speed will be small – small enough that the acceleration can be calculated and assumed constant, allowing the motion during this short interval to be determined. With the help of a computer (because intervals of 0.1 s means a lot of data points), you get a plot of time versus puck position, which shows for a puck of mass 170 g, the stopping distance is 227 m. Turns out, air resistance plays a significant role.

So what about hitting the puck around an entire hockey rink (about 180 m, in the shape of a rounded rectangle) with one shot? In this scenario, the motion of the puck can be split into two parts. The simple part is the motion along the straight edges of the rink – the wall would create a different interaction with the air, and change the drag coefficient. For a first approximation we can assume the puck follows the same calculation as above. But when the puck travels around the rounded corners of the rink, which have a radius of curvature of 8.5 m, the boundary wall will add two new forces to the calculation. First, the normal force from the wall, which pushes the puck sideways in order to get it to turn. Second, the friction between the wall and the puck.

There are also two ways a puck could travel around this bend. It could “roll” along the wall, in which case there will still need to be some type of wall frictional force that causes the puck to increase its angular velocity. In order to be completely rolling, the angular velocity of the puck would have to be equal to the linear speed of the puck multiplied by the radius of the puck (which is true for any rolling without slipping object).

The other way the puck could travel around the corners is by completely sliding without rolling. In this version, the angular velocity of the puck would stay at zero and there would just be a kinetic frictional force. Of course, the coefficient of friction between the rubber puck and the wall would likely be much higher than for the ice–rubber interaction. But it gets even more complicated.

For this wall–puck friction, the magnitude of the frictional force depends on the normal force for the wall pushing on the puck to make it turn. This normal force depends on the radius of curvature of the wall and the speed of the puck. So, again, the wall–puck frictional force depends on the speed of the puck.

It’s possible that the puck could “bounce” when it transitions from the straight part of the wall to the curved part. This would not only cause it to lose kinetic energy (and slow down), but it would also mean that it loses contact with the wall. This is a pretty tough problem, and to solve it you probably need some more experimental data on the interaction between the puck and the wall. Consider this your homework the next time you’re watching a game!