Animated Solution for Physics - Kinematics: A particle moves such that its position vector $\mathbf{r}(t)=\cos \omega t\hat{\mathbf{i}}+\sin \omega t\hat{\mathbf{j}}$, where $\omega$ is a constant and $t$ is time. Then, which of the following statements is true for the velocity $\mathbf{v}(t)$ and acceleration $\mathbf{a}(t)$ of the particle?

Unraveling Uniform Circular Motion

Imagine a particle gracefully tracing a path in the $xy$-plane. We are given its position vector as a function of time:

At first glance, this might just look like a mix of trigonometric functions, but it holds a beautiful geometric secret. If we look at the $x$ and $y$ components, we have $x=\cos (\omega t)$ and $y=\sin (\omega t)$. If we square both components and add them together, we get $x^2+y^2={\cos }^2(\omega t)+{\sin }^2(\omega t)=1$. This is the mathematical signature of a circle with a radius of 1! The particle is performing a uniform circular motion.

To understand how the particle is moving, we need to find its velocity. In kinematics, velocity is simply the rate of change of position. So, we differentiate the position vector $\mathbf{r}(t)$ with respect to time $t$.

Using the chain rule, the derivative of $\cos (\omega t)$ is $-\omega \sin (\omega t)$, and the derivative of $\sin (\omega t)$ is $\omega \cos (\omega t)$.

Now, is there a special relationship between the velocity and the position? Let's take their dot product.

Notice how the two terms are identical but have opposite signs. They perfectly cancel out to give zero! A dot product of zero mathematically proves that the velocity vector is always perpendicular to the position vector. Physically, this means the velocity is always tangential to the circular path.

Next, let's find the acceleration by differentiating the velocity vector with respect to time.

Differentiating again, another $\omega$ pops out from the chain rule.

Let's factor out the common term, $-{\omega }^2$.

Look closely at the terms inside the bracket. It is exactly our original position vector $\mathbf{r}(t)$!

This final equation is profound. The position vector $\mathbf{r}$ points radially outwards from the origin to the particle. The negative sign in our acceleration equation indicates that the acceleration vector points in the exact opposite direction—radially inwards, straight towards the origin.

This is the classic definition of centripetal acceleration. It doesn't speed up or slow down the particle; it purely acts as a turning force, constantly pulling the particle towards the center to keep it on its circular path.

Therefore, we have beautifully proven that the velocity is perpendicular to the position vector, and the acceleration is directed towards the origin.