Question

A particle is moving in the $x$-$y$ plane according to $r = b \cos \omega t \hat{i} + b \sin \omega t \hat{j}$. Where $\omega$ is constant. Which of the following statement(s) is/are true?

• A. E/ω is a constant where E is the total energy of the particle.
• B. The trajectory of the particle in x-y plane is a circle.
• C. In ax- ay plane, trajectory of the particle is an ellipse (ax, ay denotes the components of acceleration)
• D. → → a= ω2v

Answer 1

The Correct Option is A, B

We need to analyze the given position vector, determine the trajectory, velocity, acceleration, and energy of the particle to determine the correct statements.

Step 1: Determine the Trajectory

The position vector is given by: \[\vec{r} = b \cos(\omega t) \hat{i} + b \sin(\omega t) \hat{j}\] Let \(x = b \cos(\omega t)\) and \(y = b \sin(\omega t)\). Then, \[x^2 + y^2 = b^2 \cos^2(\omega t) + b^2 \sin^2(\omega t) = b^2 (\cos^2(\omega t) + \sin^2(\omega t)) = b^2\] Since \(x^2 + y^2 = b^2\), the trajectory is a circle of radius \(b\) centered at the origin.

Step 2: Calculate Velocity

The velocity vector is the time derivative of the position vector: \[\vec{v} = \frac{d\vec{r}}{dt} = -b\omega \sin(\omega t) \hat{i} + b\omega \cos(\omega t) \hat{j}\]

The speed \(v\) is the magnitude of the velocity vector: \[v = \sqrt{(-b\omega \sin(\omega t))^2 + (b\omega \cos(\omega t))^2} = \sqrt{b^2 \omega^2 (\sin^2(\omega t) + \cos^2(\omega t))} = b\omega\]

So, the speed is constant.

Step 3: Calculate Acceleration

The acceleration vector is the time derivative of the velocity vector: \[\vec{a} = \frac{d\vec{v}}{dt} = -b\omega^2 \cos(\omega t) \hat{i} - b\omega^2 \sin(\omega t) \hat{j} = -\omega^2 (b \cos(\omega t) \hat{i} + b \sin(\omega t) \hat{j}) = -\omega^2 \vec{r}\]

The magnitude of the acceleration is: \[a = |\vec{a}| = \omega^2 |\vec{r}| = \omega^2 b\]

Step 4: Calculate Energy

Since we are not given any potential energy, we can assume the total energy \(E\) is the kinetic energy: \[E = \frac{1}{2} mv^2 = \frac{1}{2} m (b\omega)^2 = \frac{1}{2} m b^2 \omega^2\]

Therefore, \[\frac{E}{\omega} = \frac{\frac{1}{2} m b^2 \omega^2}{\omega} = \frac{1}{2} m b^2 \omega\]

Since m and b are constant as well as w, \(\frac{E}{\omega}\) is proportional to omega. Hence first statement seems incorrect. Recalculating the statement one. Kinetic energy = 1/2 mv^2. Since speed = bw. KE is 1/2 * m * b^2 * w^2. \(\frac{E}{w} = 1/2 m b^2 w \) Since b is constant, w is constant then this value is constant.

 

Step 5: Analyze the Statements

1. E/\(\omega\) is a constant where E is the total energy of the particle: This statement is TRUE as \(E/\omega = \frac{1}{2} m b^2 \omega \) is a constant.
2. The trajectory of the particle in x-y plane is a circle: This statement is TRUE as \(x^2 + y^2 = b^2\).
3. In a_x- a_y plane, trajectory of the particle is an ellipse (a_x, a_y denotes the components of acceleration): This statement is FALSE. The trajectory is a circle.
4. \(\vec{a} = \omega^2 \vec{v}\): This statement is FALSE. \(\vec{a} = -\omega^2 \vec{r}\)

Conclusion

The correct statements are:

• E/\(\omega\) is a constant where E is the total energy of the particle.
• The trajectory of the particle in x-y plane is a circle.