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4. Two Dimensional Kinematics

We frequently encounter motion in more than one dimension. A common example studied in introductory physics courses is the projectile motion -- the motion of an object thrown up in the air at some angle \(\theta\). As such, the object (projectile) has both a vertical and a horizontal displacement.

Components of displacement

\[ \Delta x = x - x_0 \]
\[ \Delta y = y - y_0 \]

where \(x_0\) is the horizontal position at \(t=0\), and \(y_0\) is the vertical position at \(t=0\). It should be understood that \(x\) and \(y\) are functions of \(t\).

4.1 Projectile Motion

In the two different directions, there are two values of acceleration. The pull due to Earth's gravity acts only in the vertical direction; as such, in the vertical direction (\(y-\) axis) the acceleration has a magnitude of about \(g=9.8\ m/s^2\).

This idea can be represented in terms of two-dimensional vectors:

\[\begin{align} a_x &= 0 \\ a_y &= -g \end{align}\]

where the negative sign indicates that the acceleration in pointing downward towards the center of the Earth as conventionally we choose up as the positive direction on the y-axis.

The acceleration vector therefore is:

\[ \vec{a} = a_x\ \hat{i} + a_y\ \hat{j} = -g\ \hat{j} \]

Mathematically, it is somewhat easier to consider the \(x\) and \(y\) components of the motion separately. In both directions the motion has a constant acceleration, and we can make use of the 1D constant acceleration equations of motion that we have derived before.

Animation of 2d projectile motion

Use slider below to change the value of the veritical acceleration.

4.1.1 Level ground

The projectile is launched from a level ground at an angle of \(\theta_0\) similar to the animation above. The horizontal \(x\) and vertical \(y\) components of various quantities are as follows:

\[ \begin{align} v_{0x} &= v_0 \cos \theta_0 \\ a_x &= 0 \\ x_0 &= 0 \\ x_f &= R \ (\rm range) \end{align} \]
\[ \begin{align} v_{0y} &= v_0 \sin \theta_0 \\ a_y &= -g \\ y_0 &= 0 \\ y_f &=0 \ ({\rm ground}) \end{align} \]

Each of the two components can be studied individually using the 1D equations of motion. For instance, let us use the vertical component to find the time of flight of the projectile:

Useful quantities

When \(t=T_{\rm tof}\), we know that \(y_f = y_0 = 0\) as the projectile comes back to the level ground. So, using one of the equations of motion for the vertical direction:

\[ \begin{align} y - y_0 &= v_{0y} t + \frac{1}{2} a_y t^2 \\ 0 - 0 &= (v_0 \sin \theta_0) T_{\rm tof} + \frac{1}{2}(-g) T_{\rm tof}^2 \\ \end{align} \]

so that

\[ T_{\rm tof} \left( v_0 \sin \theta_0 - \frac{g T_{\rm Tof}}{2} \right) = 0 \\ \]

This equation has two solutions.

\[{\rm Either \quad} T_{\rm tof} = 0, \quad {\rm or } \quad T_{\rm tof} = \frac{2 v_0 \sin \theta_0}{g} \]

Think about which one of them is the correct expression for \(T_{\rm tof}\).

The horizontal range can be found by considering the \(x-\)component with the final time as the time of flight. Recall also that \(a_x = 0\).

\[ \begin{align} x - x_0 &= v_{0x} t + \frac{1}{2} a_x t^2 \\ R - 0 &= v_0 \cos \theta_0 T_{\rm tof} + 0 \\ R &= v_0 \cos \theta_0 \left( \frac{ 2 v_0 \sin\theta_0}{g} \right) \\ R &= \frac{ 2 v_0^2 \sin \theta_0 \cos \theta_0 }{g} \end{align} \]

Further using the trigonometric identity \(\sin(2\theta) = 2 \sin \theta \cos \theta\), we obtain:

\[ R = \frac{v_0^2 \sin(2\theta_0)}{g} \]

To find the maximum height one approach is to use the half of the time of flight. Another approach is to use the following equation of motion:

\[ v_y^2 = v_{0y}^2 + 2 a_y (y-y_0) \]

and realize that at the top (\(y-y_0=H\)), \(v_y =0\), so that:

\[ 0 = (v_0 \sin \theta_0)^2 + 2 (-g) H \]
\[ H = \frac{v_0^2 \sin^2 \theta_0}{2 g} \]

where we have also used the fact that \(a_y = - g\) (acceleration due to gravity).

What initial angle \(\theta_0\) gives the maximum horizontal range?

Hint: This question can be answered using the range formula

\[ R = \frac{v_0^2 \sin(2\theta_0)}{g}\]

For constant vlues of \(v_0\) and \(g\), what value of \(\theta_0\) gives the maximum value for \(\sin(2\theta_0)\)?

4.2 Uniform Circular Motion

Consider a motion of a object along a circular path of radius \(r\) at a constant speed of \(v\).

Let the center of the circular path be the origin of the co-ordinate system such that the line connecting the object and the origin makes an angle \(\theta\) with the positive \(x\)-axis.

As the object moves in the circle, this angle \(\theta\) is constantly changing; for constant speed, we can write: \(\theta = \omega t\) where \(\omega\) is the angular speed in \(rad/s\).

Vectors in uniform circular motion

The position vector of the particle in circular motion therefore is:

\[\vec{r}(t) = (r \cos \omega t) \hat{i} + (r \sin \omega t) \hat{j} \]

The velocity vector can be obtained by taking the derivative of the position vector:

\[ \vec{v}(t) = \frac{d \vec{r}(t)}{dt} = -r \omega \sin (\omega t) \hat{i} + r \omega \cos (\omega t) \hat{j} \]

The acceleration vector can be obtained by taking the derivative of the velocity vector:

\[ \vec{a}(t) = \frac{d \vec{v}(t)}{dt} = -r \omega^2 \cos(\omega t) \hat{i} - r \omega^2 \sin (\omega t) \hat{j} \]

From the expressions for the position and acceleration vectors for an object in a uniform circular motion, we can see that \(\vec{a}(t) = - \omega^2 \vec{r}(t)\). That is the acceleration vector is always pointing towards the center.

4.2.1 Formulas for uniform circular motion

  • Centripetal acceleration, \(a_c = |\vec{a}_c| = \frac{v^2}{r}\).
  • Speed, \(v = r \omega\), where \(\omega\) is the angular speed in \(rad/s\).
  • Angular speed, \(\omega = \frac{2 \pi}{T}\), whee \(T\) is the time period in \(s\).
  • Also, \(\omega = 2 \pi f\), where \(f\) is the frequency in \(Hz\) or \(s^{-1}\).
  • Also, \(a_c = r \omega^2\)