# HKDSE Physics and M2 (Part II)

Continuing from part I, we would now look into projectile motion and uniform circular motion.

I would highly recommend checking out this video on vectors before reading the post. Having a general idea of what vectors are would be extremely helpful.

## Motion on the 2D plane

To study motion on the 2D plane, we need the idea of curves. The trajectory of a moving particle naturally forms a curve as time varies.

Mathematically, we model a curve as a function from \(\mathbb{R}\), the real numbers, to \(\mathbb{R}^2\), the cartesian plane. Here are some examples below. The playback speed of the gifs may vary.

### Straight line

The function \(r(t) = (t, 0), \text{ for } 0 < t < 1\) defines the curve below. You could imagine it as a ball moving 1 unit on the x-axis from \(t=0\) to \(t=1\). Without doing any mathematics, you could intuitively see that the velocity is going to be constant and so acceleration would be 0.

Mathematically, we can take the derivative of \(r(t)\) by taking the derivative of its components. So we would have \(r'(t) = (1,0)\). This would be the velocity of the ball, a constant, unit vector pointing towards the x-axis. The acceleration, as you can guess, would be \(r''(t) = (0,0)\) which is the 0 vector.

### Parabola

The function \(r(t) = (3t, 4t - \frac{9.81}{2} t^2) \text{ for } 0 < t < 1\) defines the curve below. You could imagine it as throwing a ball at the origin under the effect of gravity. Could you try to understand this motion by considering the coordinates separately and using the equations we developed in part I?

Intuitively, we know that the velocity would have the same x-component for all time t and that the acceleration would be a constant vector pointing downwards.

Mathematically, we have \(r'(t) = (3, 4 - 9.81t)\) and \(r''(t) = (0, -9.81)\). In the gif below, the green vector is the velocity and the blue vector is the acceleration. Both vectors’ magnitude are scaled down by a factor of 1/10. You could play around with the graph here.

### Circle

The function \(r(t) = (\cos t, \sin t) \text{ for } 0 < t < 2 \pi\) defines the curve below. You can imagine as a ball uniformly rotating around the origin with radius 1.

Intuitively, we know that the velocity would be the tangent vector to the circle. The magnitude would be constant (1) as the motion is uniform. Acceleration would also be constant and pointing towards the origin.

Mathematically, we have \(r'(t) = (-\sin t, \cos t)\) and \(r''(t) = (-\cos t, -\sin t)\) which aligns with our intuition. Once again the green vector is the velocity and the blue vector is the acceleration. You can play around with the graph here

As you can see the amount of behaviour we can model with curves (the explicit construction of the \(r(t)\) function is called curve parameterization) is highly unconstrained! It is powerful enough to describe a far wider range of curves than just plots of \(y=f(x)\) (which one can imagine as plotting \(r(t) = (t, f(t)) \; \forall t \in \mathbb{R}\)). There are other ways of constructing curves such as using level sets. For those interested in this fascinating topic, you could read chapter 2 of this set of lecture notes from CUHK.

## Projectile motion

Similar to part I, the crucial assumption in DSE projectile motion is that the only force exerted on the particle is the gravitational force. So once again we have

\[\begin{align*} r''(t) &= (0, -g) \\ r'(t) &= (C_1^*, -gt + C_1) \\ r(t) &= (C_1^* t + C_2^*, -\frac{1}{2}gt^2 + C_1 t + C_2 ) \end{align*}\]for some constants \(C_1^*, C_2^*, C_1, C_2\) by repeated “integration”. Similar to part I we could find those constants in terms of initial velocities / displacement. As such most properties of projectile motion could be analysed by splitting into x and y-axis.

Another way of looking at it courtesy of Henry Yip would be to consider

\[r(t) = (C_2^*, C_2) + (C_1^*, C_1) t + (0, -\frac{1}{2}g) t^2\]which tells you that for small \(t\), \(r(t)\) looks like a straight line starting from initial displacement \((C_2^*, C_2)\) with the direction of initial velocity \((C_1^*,C_1)\). Gradually the quadratic term dominates and we get the parabolic shape. This idea is similar to taylor expansions.

Perhaps, then, the most interesting aspect about projectile motion is the conservation of energy. Why is it that energy is still conserved when we use the magnitude of the velocity vector in kinetic energy (instead of one dimensional velocity)? How does the formalism developed in part I relate to the 2 dimensional case? Let’s make some definitions first.

Let \(r(t) = (x(t), y(t))\). So \(x(t), y(t)\) are the x and y components of \(r(t)\) respectively. As such we have \(r'(t) = (x'(t), y'(t))\) and \(r''(t) = (x''(t), y''(t))\). Now we have

\[\begin{align*} \text{Kinetic energy } &= T = \frac{1}{2} m (x'(t)^2 + y'(t)^2) \\ \text{Potential energy } &= V = mgy(t) \\ \end{align*}\]So we have

\[\begin{align*} &\phantom{= } T + V \\ &= \frac{1}{2} m (x'(t)^2 + y'(t)^2) + mgy(t) \\ &= \frac{1}{2} m ({C_1^*}^2 + (-gt + C_1)^2) +mg(-\frac{1}{2}gt^2 + C_1 t + C_2 ) \\ &= m\bigg[\frac{1}{2} {C_1^*}^2 + \frac{1}{2} g^2 t^2 - gtC_1 + \frac{1}{2}C_1^2 - \frac{1}{2} g^2 t^2 + gC_1 t + gC_2\bigg] \\ &= \frac{1}{2} m ({C_1^*}^2 + C_1^2 ) + mg C_2 \end{align*}\]which is the total energy at initial time.

A more proper way of doing this would involve multivariable calculus. Again refer to this set of lecture notes for a more general analysis on conservative forces.

## Uniform circular motion

Let’s think about a ball uniformly rotating around the origin. We know that two variables completely determine its behaviour, its radius and its velocity. As such we can parameterize \(r(t) = (R \cos (kt), R \sin (kt))\) where \(R\) is the radius and \(k\) is some variable that as it turns out is related to angular velocity.

To intuitively see why \(k\) is related to angular velocity: Consider how \(r(t) = (\cos(t), \sin(t)), 0<t<2 \pi\) is one full anticlockwise rotation around the unit circle, but \(r(t) = (\cos(2t), \sin(2t)), 0<t<\pi\) is the same full anticlockwise rotation in half the time. We doubled \(k\) and the time taken is halved. Could you guess a relationship between \(k\) and angular velocity before we do the maths?

Let’s find out the velocity and the acceleration. We have

\[\begin{align*} r(t) &= (R \cos (kt), R \sin (kt)) \\ r'(t) &= (-Rk \sin (kt), Rk \cos (kt))\\ r''(t) &= (-Rk^2 \cos (kt), -Rk^2 \sin (kt)) \end{align*}\]These formulae immediately tell us all we know about uniform circular motion!

Firstly, \(r(t) \perp r'(t) \perp r''(t)\) from simple coordinate geometry (or you could use the dot product if you are familiar with linear algebra).

Secondly, the magnitude of the velocity is \(\sqrt{R^2k^2 ( \sin^2 (kt) + \cos^2 (kt))} = Rk\). So we now know \(v = Rk\).

What about angular velocity? We see that for a full anticlockwise rotation to take place, \(t\) needs to go from \(0\) to \(2\pi/k\). The total angular change would be \(2 \pi\). As such the angular velocity is \(\frac{2 \pi k}{2 \pi} = k\). So \(k\) is the angular velocity!

Finally, \(r''(t) = -k^2 r(t)\), so \(a = k^2 R\)!

As such we also have \(a = v^2 / R\)