Question

The position vector of the particle is $\vec{r}(t)=a\cos\omega t\,\hat{i}+a\sin\omega t\,\hat{j}$, where $a$ and $\omega$ are real constants of suitable dimensions. The acceleration is

• A. perpendicular to the velocity
• B. parallel to the velocity
• C. directed away from the origin
• D. perpendicular to the position vector
• E. always along the direction of $\hat{i}$

Hint:

In any motion where the magnitude of velocity is constant (like uniform circular motion), the acceleration is always perpendicular to the velocity.

Answer 1

The Correct Option is A

Concept:
The velocity vector is the first derivative of the position vector, and acceleration is the second derivative. [itemsep=8pt]
• Velocity ($\vec{v}$): $\frac{d\vec{r}}{dt} = -a\omega \sin \omega t \, \hat{i} + a\omega \cos \omega t \, \hat{j}$
• Acceleration ($\vec{a}$): $\frac{d\vec{v}}{dt} = -a\omega^2 \cos \omega t \, \hat{i} - a\omega^2 \sin \omega t \, \hat{j} = -\omega^2 \vec{r}$

Step 1: Check the relationship between $\vec{a}$ and $\vec{v}$.
To determine the geometric relationship, we compute the dot product $\vec{a} \cdot \vec{v}$: \[ \vec{a} \cdot \vec{v} = (-a\omega^2 \cos \omega t)(-a\omega \sin \omega t) + (-a\omega^2 \sin \omega t)(a\omega \cos \omega t) \] \[ \vec{a} \cdot \vec{v} = a^2\omega^3 \sin \omega t \cos \omega t - a^2\omega^3 \sin \omega t \cos \omega t = 0 \]

Step 2: Conclusion.
Since the dot product is zero, the acceleration vector is always perpendicular to the velocity vector.
[8pt] This is a characteristic of uniform circular motion where the acceleration is radial and velocity is tangential.