Chenying Liu

Are you a fan of the cartoon Tom and Jerry? Have you ever watched the episode ‘The Wacky World of Sports?’ The most famous human (animal?) drama of athletic competition! Tom Cat and Jerry Mouse compete for the gold medal in a decathlon. Tennis, pole-vault, martial arts, 100-yard dash, swimming, weightlifting, long jump, golf, and discus throw. A total of nine track and field events lasting the whole day. Although Tom was never expected to win any competition, the thrilling championship remains technically unresolved because we of course need ten events for a true decathlon. This is the final endgame…

In what event will they compete for the final challenge? Well, how about something they enjoy every day? That is, Chase and Escape! The rules are simple: Tom chases Jerry in an empty stadium, which is in the shape of a circle; the stadium is closed so no one can leave; and they are only allowed to use the same constant speed. If Tom catches Jerry, Tom wins; otherwise, he will have to suffer the agony of defeat once again.

Rules

Before the game starts, Tom and Jerry are airdropped into the stadium and thus, their initial positions are considered to be random (of course we assume they don’t start at the same spot). In addition, due to their nature as a cat and a mouse, we assume it takes no time to respond to each other’s actions. 

Ready! Set! Go! The thrilling competition begins…

Tom’s Strategy

First let’s look at Tom. One tactic could be to head straight to the centre of the stadium instead of going towards Jerry’s location. But, is this a good idea?

We can think about it as follows. In Stage 1, by going to the centre (labelled as a black point in the diagram below), Tom can take the initiative.  If Jerry was near to the centre, he would escape to the edge to avoid being caught by Tom directly. If not, Jerry wouldn’t come to the centre either as he can see Tom is already heading in that direction. No matter what happens, this means Tom will be closer to the centre than Jerry. But, what advantage will this bring to Tom? The key point is that no matter where Jerry goes, it turns out that Tom can find a way to stay on the straight line between Jerry and the central point as shown in stage 3 below. We will see why this gives him an advantage shortly, but first let’s look at how Tom is able to ensure this will happen. 

Let’s consider Stage 2 in the diagram above during which Tom is closer to the centre than Jerry but they are not yet on the same line with the central point. To understand Tom’s strategy, we need to introduce the concept of angular velocity.

Normally, velocity along a straight line, which we call linear velocity or v, is measured in units of length l divided by time t. In modern physics, that’s usually meters per second (m/s). If we look at the diagram below we can see that when Tom runs 6 meters on the straight line in 2 seconds, his linear velocity is 6 divided by 2, which equals 3 meters per second. Hopefully, this is something that you are already familiar with. However, things are quite different in the rotational world. 

If Tom runs along the circumference of a circle as illustrated above, his velocity is always perpendicular (at a right-angle) to the radius. In such a circular motion, we introduce angular velocity and describe it in terms of the central angle divided by time t. It’s units can be degrees per second (s^-1) or radians per second (rad/s), where we multiply by π/180 to convert from degrees to radians. For example, 30 degrees equals 30π/180= π/6 radians.

Now, let’s look again at the diagram labelled ‘Angular Velocity’ above.  As Tom runs along the circle from point A to point B, the angle of the arc he covers is 90º and let’s say the time he takes is 4 seconds. Then his angular velocity is 90 divided by 4, equal to 22.5 degrees per second, or π/8 radians per second. The path he follows is the arc shown  in red.

Now, you may be wondering if there is a relationship between linear and angular velocity? And the answer is yes! When Tom runs on the circle, the total distance he has covered is the arc length and so therefore the corresponding linear velocity is the arc length divided by time. The arc length s can be calculated by the formula s = r α, where s = arc length, r = radius, α = central angle in radians as marked in the picture above. The arc length is also equal to the corresponding linear velocity v, multiplied by time: s = v t. Since we already know the angular velocity ω = α / t, substituting this into the formula s = v t gives s = v a / ω. Compare this with the formula s = r a. Then alpha is eliminated and the result gives v = r ω, which means the linear velocity is equal to the radius multiplied by the angular velocity. 

In our calculations above we used the fact that the linear velocity v, is the one perpendicular to the radius. If its angle with the radius is not 90º, then we instead need to use its component perpendicular to the direction of the radius. The diagram below illustrates how this works. If we label the angle between the velocity and the radius as θ, we can then decompose the velocity into two directions. The one which is perpendicular to the radius (and thus is the one used to calculate angular velocity) is v sinθ. We therefore, have the final expression of angular velocity as ω = v sinθ / r. You may also have noticed a second velocity component labelled in the diagram which is the direction parallel to the radius with value v cosθ. Take note as we’ll use this later.

Now that we have an understanding of angular velocity, let’s go back to Stage 2 in Tom’s strategy.

Let’s call the distance from the centre to Tom r₁ and that from the centre to Jerry r₂. According to the rules of the Chase and Escape game, they are only allowed to use the same constant speed v.  Armed with our new knowledge we can calculate Tom’s angular velocity ω₁ and Jerry’s angular velocity ω₂ as 

where θ₁ and θ₂ are the angles between their velocities and the lines r₁ and r₂. We also have their velocities in the direction parallel to the radius as |v| cosθ₁ and |v| cosθ₂. As mentioned before, keep these in mind as they are really important!

Now, if Tom just keeps θ₁ as 90º, then his angular velocity ω₁ is always |v| / r₁. When it comes to Jerry, since -1 ≤ sinθ₂ ≤ 1, the maximum of his angular velocity ω₂ is |v| / r₂. Now, remember that Tom is closer to the centre, so we have that r₁ < r₂ and therefore Tom’s angular velocity will be larger than that of Jerry (since it’s larger than Jerry’s maximum). This indicates that in the same time period, Tom can cover a larger central angle than Jerry. Thus, no matter how far Tom has lagged behind, eventually he will catch up to make their real-time positions and the centre collinear (all lying along the same line as shown in stage 3). Then Tom will be between Jerry and the centre and Jerry is prevented from going to the centre (since if he goes, he will definitely meet Tom on the way there).

Now, we can move on to Stage 3. 

To maintain collinearity, Tom only needs θ₁ to be smaller than θ₂, which brings another advantage. Their velocities along the radius are |v| cosθ₁ and |v| cosθ₂ (as we have discussed before). Here, when 0º ≤ θ ≤ 180º, a smaller gives a larger value of cos, and hence Tom will possess a higher velocity along the radius, which means he can constantly narrow down the distance to Jerry. As he does so, they are both also approaching the edge, and so Tom is quite confident that he will at the very least catch Jerry there (if not before). It seems that Tom’s success is only a matter of time…

If only it were so easy! Whilst we know Tom is approaching the edge,  does this guarantee that he will in fact finally reach the edge? The answer is perhaps uncertain and we can see why by considering the following example.

A man has signed up for a 10-km walking race. His full speed is 5 km per hour, which means it only takes 2 hours to finish at this rate. However, he is reducing his speed by half every one hour. This means, he can walk 5 km in the first hour, 2.5 km in the second hour, and 1.25 km in the third hour… Although he keeps approaching the end, he can never reach it as he can only move half of the distance left each time. So, perhaps Tom is in a similar situation to the man in our race?

In the meantime, let’s take a look at Jerry. What will our little Mouse do? Does he have any winning strategy to defeat his arch nemesis? See if you can work it out and then check back to see the answer in part 2!