Animated Solution for Physics - Kinematics: In three dimensional system, the position coordinates of a particle (in motion) are given below $x=a\cos \omega t,y=a\sin \omega t,z=a\omega t$ The velocity of particle will be

Imagine a particle moving through three-dimensional space. Its position isn't just a simple line or a flat curve; it's a complex dance defined by three coordinates: $x$, $y$, and $z$.

The problem gives us these coordinates as functions of time: $x=a\cos \omega t$, $y=a\sin \omega t$, and $z=a\omega t$.

If we look closely at the $x$ and $y$ components, they form the parametric equations of a circle. This means that if we were to look down at the particle from above (the $xy$-plane), it would appear to be moving in a perfect circle of radius $a$.

However, the $z$ component is $a\omega t$, which means the particle is constantly moving upwards at a steady rate. Combining this circular motion with the upward motion, the particle traces out a helix—much like a spiral staircase!

To find the velocity of the particle, we need to translate its position into a mathematical vector. The position vector $\vec{r}$ is simply the combination of its coordinates:

Substituting our given functions, we get:

In kinematics, velocity is the time derivative of position. So, we must differentiate our position vector with respect to time $t$:

Let's differentiate term by term. The derivative of $\cos \omega t$ is $-\omega \sin \omega t$, the derivative of $\sin \omega t$ is $\omega \cos \omega t$, and the derivative of $t$ is just $1$. Applying this, we find the velocity vector:

Now, take a look at the options provided in the question. They are all scalar values (like $\sqrt{2}a\omega$). This is a crucial hint!

Even though the question asks for "velocity", the options indicate that it actually wants the magnitude of the velocity, which is the speed.

The magnitude of any 3D vector is found using the 3D Pythagorean theorem:

Let's plug our velocity components into the magnitude formula:

When we square the negative $x$-component, the negative sign vanishes. Expanding the squares gives us:

Notice that the term $a^2{\omega }^2$ is common to all three parts under the square root. Let's factor it out from the first two terms:

Here, we encounter the most famous trigonometric identity: ${\sin }^2\theta +{\cos }^2\theta =1$. Substituting this in, the expression simplifies beautifully:

Finally, pulling the perfect squares out of the square root, we arrive at our answer:

This result is fascinating. It tells us that even though the particle's velocity vector is constantly changing direction as it spirals upwards, its speed remains absolutely constant!