Uniform circular motion: a kinematics problem


Posted by Diego Assencio on 2013.11.16 under Physics (Kinematics)

Consider an object traveling along a circle of radius $r$ on the $(x,y)$ plane (see figure 1). The circle is centered at the origin $O$ of the plane. The object travels with constant angular velocity $\omega$, meaning the angle $\theta(t)$ formed by its path (which is an arc) with the positive $x$ axis at time $t$ is given by: $$ \theta(t) = \theta_0 + \omega t $$ where $\theta_0$ is the value of $\theta(t)$ at $t = 0$. We will assume here that $\theta_0 = 0$. In other words, $\theta(t=0) = 0$ so the object is at the position $(x_0,y_0) = (r,0)$ at that time, meaning it initially lies at the positive $x$ axis at a distance $r$ from the origin $O$.

Fig. 1: An object traveling along a circle of radius $r$ with constant angular velocity $\omega$.

The equations of motion for the object are given by (see figure 1): $$ \begin{eqnarray} x(t) &=& r\cos\theta(t) &=& r\cos(\omega t) \label{post_83aa8ba50acf8f902bd6850862846e17_eq_motion_x}\\[5pt] y(t) &=& r\sin\theta(t) &=& r\sin(\omega t) \label{post_83aa8ba50acf8f902bd6850862846e17_eq_motion_y} \end{eqnarray} $$

The velocities and accelerations along the $x$ and $y$ directions can be obtained directly from equations \eqref{post_83aa8ba50acf8f902bd6850862846e17_eq_motion_x} and \eqref{post_83aa8ba50acf8f902bd6850862846e17_eq_motion_y} (below the notation $\dot{q}$ is used to represent $dq/dt$): $$ \begin{eqnarray} v_x(t) &=& \dot{x}(t) &=& -\omega r \sin(\omega t) \quad\quad & a_x(t) &=& \dot{v}_x(t) &=& -\omega^2 r \cos(\omega t) \label{post_83aa8ba50acf8f902bd6850862846e17_vel_accel_x}\\[5pt] v_y(t) &=& \dot{y}(t) &=& \omega r \cos(\omega t) \quad\quad & a_y(t) &=& \dot{v}_y(t) &=& -\omega^2 r \sin(\omega t) \label{post_83aa8ba50acf8f902bd6850862846e17_vel_accel_y} \end{eqnarray} $$ Here is where the interesting physics starts. Let the position vector ${\bf x}(t)$ be the vector connecting the origin $O$ to the mass $m$ as in figure 1. In other words: $$ {\bf x}(t) := (x(t),y(t)) = r(\cos(\omega t), \sin(\omega t)) \label{post_83aa8ba50acf8f902bd6850862846e17_eq_motion_vec} $$ Then writing ${\bf v}(t) = (v_x(t), v_y(t))$ and ${\bf a}(t) = (a_x(t), a_y(t))$ for the velocity and acceleration vectors at time $t$ respectively we obtain the following: $$ {\bf v}(t) = \omega r (-\sin(\omega t), \cos(\omega t)) \label{post_83aa8ba50acf8f902bd6850862846e17_vel_vec} $$ $$ {\bf a}(t) = -\omega^2 r (\cos(\omega t), \sin(\omega t)) = -\omega^2 {\bf x}(t) \label{post_83aa8ba50acf8f902bd6850862846e17_accel_vec} $$ where equations equations \eqref{post_83aa8ba50acf8f902bd6850862846e17_eq_motion_x}, \eqref{post_83aa8ba50acf8f902bd6850862846e17_eq_motion_y}, \eqref{post_83aa8ba50acf8f902bd6850862846e17_vel_accel_x} and \eqref{post_83aa8ba50acf8f902bd6850862846e17_vel_accel_y} were used. In other words, since ${\bf x}(t)$ always points away from the center $O$ of the circle, then the acceleration ${\bf a}(t)$, called centripetal acceleration, points always towards the center of the circle! To gain more intuition about why this is so, notice first that the velocity vector ${\bf v}(t)$ is perpendicular to the position vector ${\bf x}(t)$ at all times: $$ \begin{eqnarray} {\bf v}(t)\cdot{\bf x}(t) &=& x(t)v_x(t) + y(t)v_y(t) \nonumber\\[5pt] &=& \omega r^2\left[\cos(\omega t)(-\sin(\omega t)) + \sin(\omega t)\cos(\omega t)\right] \nonumber\\[5pt] &=& 0 \end{eqnarray} $$ where equations \eqref{post_83aa8ba50acf8f902bd6850862846e17_eq_motion_vec} and \eqref{post_83aa8ba50acf8f902bd6850862846e17_vel_vec} were used. So for circular motion, the velocity of the object is always tangential to the circle. This is true even if the angular velocity $\omega$ is not constant over time (I strongly recommend you check this fact on your own by computing the derivative of ${\bf x}(t) = (r\cos\theta(t),r\sin\theta(t))$ with respect to time for an arbitrary $\theta(t)$ and taking the dot product of ${\bf x}(t)$ and ${\bf v}(t)$). Since in our case the motion is circular with constant angular velocity $\omega$, the magnitude of the velocity must be always constant. In fact: $$ v := \|{\bf v}(t)\| = \sqrt{v_x^2(t) + v_y^2(t)} = \sqrt{\omega^2 r^2 \sin^2(\omega t) + \omega^2 r^2\cos^2(\omega t)} $$ and since $\sin^2\alpha + \cos^2\alpha = 1$ for any real value $\alpha$, we get: $$ \boxed{\phantom{\|}v = \omega r\phantom{\|}} \label{post_83aa8ba50acf8f902bd6850862846e17_mag_v} $$ As shown above, $v$ denotes the magnitude of ${\bf v}(t)$. The value of $v$ is constant over time since both $\omega$ and $r$ are constants, so the direction of ${\bf v}(t)$ changes but its magnitude remains always the same.

The velocity ${\bf v}(t)$ and the acceleration ${\bf a}(t)$ are also always perpendicular to each other. To see this, notice that from equations \eqref{post_83aa8ba50acf8f902bd6850862846e17_vel_accel_x} and \eqref{post_83aa8ba50acf8f902bd6850862846e17_vel_accel_y} we have that: $$ \begin{eqnarray} {\bf v}(t)\cdot{\bf a}(t) &=& v_x(t)a_x(t) + v_y(t)a_y(t) \nonumber\\[5pt] &=& \omega^3 r^2\left[\sin(\omega t)\cos(\omega t) - \cos(\omega t)\sin(\omega t)\right] \nonumber\\[5pt] &=& 0 \end{eqnarray} $$ The effect of the acceleration ${\bf a}(t)$ is then to constantly change the direction of the velocity ${\bf v}(t)$ without changing its magnitude. To better understand this, let's compute how the magnitude of a general ${\bf v}(t)$ (for any kind of motion) changes over time: $$ \begin{eqnarray} \displaystyle{\frac{d\|{\bf v}(t)\|}{dt}} &=& \displaystyle{\frac{d}{dt}}\sqrt{v_x^2(t) + v_y^2(t)} \nonumber\\[5pt] &=& \displaystyle\frac{2v_x(t)a_x(t) + 2v_y(t)a_y(t)}{2\sqrt{v_x^2(t) + v_y^2(t)}} \nonumber\\[5pt] &=& \displaystyle\frac{{\bf v}(t)\cdot{\bf a}(t)}{\|{\bf v}(t)\|} \label{post_83aa8ba50acf8f902bd6850862846e17_dv_dt} \end{eqnarray} $$ so when ${\bf a}(t)$ is perpendicular to ${\bf v}(t)$, the magnitude of ${\bf v}(t)$ remains constant (but its direction changes if the acceleration is not zero!). Since we made no assumptions on ${\bf v}(t)$ and ${\bf a}(t)$ when deriving equation \eqref{post_83aa8ba50acf8f902bd6850862846e17_dv_dt}, it must be valid for any kind of motion.

Given that the object travels always in a circle of radius $r$, then $\|{\bf x}(t)\| = r$ for all $t$, so from equation \eqref{post_83aa8ba50acf8f902bd6850862846e17_accel_vec} the magnitude $a$ of the acceleration is also constant and given by: $$ a := \|{\bf a}(t)\| = \omega^2 \|{\bf x}(t)\| \Longrightarrow \boxed{ a = \omega^2 r = \displaystyle\frac{v^2}{r} } $$ where equation \eqref{post_83aa8ba50acf8f902bd6850862846e17_mag_v} was used.

Figure 2 shows the velocities and accelerations when $\theta = 0,\pi/2,\pi$ and $3\pi/2$. To verify the quantities shown below, use these values of $\theta = \theta(t) = \omega t$ on equations \eqref{post_83aa8ba50acf8f902bd6850862846e17_vel_vec} and \eqref{post_83aa8ba50acf8f902bd6850862846e17_accel_vec}.

Fig. 2: Accelerations (blue) and velocities (red) when $\theta= 0,\pi/2,\pi$ and $3\pi/2$. Notice how the acceleration always points to the center of the circle while the velocity is always tangential to the circle.

To sum up, whenever an object moves along a circular path with constant angular velocity, its acceleration points always to the center of the circle while its velocity is always tangential to the circle. Notice that we have not discussed what causes the motion of the object; we just assumed it moves along a circle while ignoring what makes it do so (in other words, this is a kinematics problem).

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