Problem 10 - Classical Dynamics of Particles and Systems

Question

A particle moves in a plane elliptical orbit described by the position vector $$\mathbf{r}=2 b \sin \omega t \mathbf{i}+b \cos \omega t \mathbf{j}$$ (a) Find \(\mathbf{v}, \mathbf{a},\) and the particle speed. (b) What is the angle between \(\mathbf{v}\) and a at time \(t=\pi / 2 \omega ?\)

Step-by-Step Solution

Answer

The velocity vector is \( \mathbf{v} = 2b\omega\cos(\omega t) \mathbf{i} - b\omega\sin(\omega t) \mathbf{j} \), and the acceleration vector is \( \mathbf{a} = -2b\omega^2\sin(\omega t) \mathbf{i} - b\omega^2\cos(\omega t) \mathbf{j} \). The particle speed is \( speed = \omega b \sqrt{4\cos^2(\omega t) + \sin^2(\omega t)} \). The angle between the velocity and acceleration vectors at \( t = \frac{\pi}{2\omega} \) is \( \theta = 90^{\circ} \) or \( \theta = \frac{\pi}{2} \) radians.

Step 1: Calculate the velocity vector \( \mathbf{v} \)

The velocity vector can be calculated by finding the first derivative of the position vector with respect to time: $$\mathbf{v} = \frac{d\mathbf{r}}{dt}$$ For the given position vector, $$\mathbf{r} = 2b\sin(\omega t) \mathbf{i} + b\cos(\omega t) \mathbf{j}$$ Taking the derivative with respect to time, we obtain: $$\mathbf{v} = \frac{d}{dt} \left( 2b\sin(\omega t) \mathbf{i} + b\cos(\omega t) \mathbf{j} \right)$$ Using the chain rule, the derivative becomes: $$\mathbf{v} = 2b\omega\cos(\omega t) \mathbf{i} - b\omega\sin(\omega t) \mathbf{j}$$

Step 2: Calculate the acceleration vector \( \mathbf{a} \)

Similarly, the acceleration vector can be calculated by finding the second derivative of the position vector with respect to time: $$\mathbf{a} = \frac{d^2\mathbf{r}}{dt^2}$$ Taking the second derivative of the position vector, we get: $$\mathbf{a} = \frac{d}{dt} \left( 2b\omega\cos(\omega t) \mathbf{i} - b\omega\sin(\omega t) \mathbf{j} \right)$$ Using the chain rule again, the derivative becomes: $$\mathbf{a} = -2b\omega^2\sin(\omega t) \mathbf{i} - b\omega^2\cos(\omega t) \mathbf{j}$$

Step 3: Calculate the particle speed

The particle speed is the magnitude of the velocity vector. We can calculate the magnitude as follows: $$speed = |\mathbf{v}| = \sqrt{(2b\omega\cos(\omega t))^2 + (-b\omega\sin(\omega t))^2}$$ Simplifying, we get: $$speed = |\mathbf{v}| = \sqrt{4b^2\omega^2\cos^2(\omega t) + b^2\omega^2\sin^2(\omega t)}$$ Factor out constant terms: $$speed = |\mathbf{v}| = \omega b \sqrt{4\cos^2(\omega t) + \sin^2(\omega t)}$$

Step 4: Determine the angle between \( \mathbf{v} \) and \( \mathbf{a} \) at \( t=\frac{\pi}{2\omega} \)

To find the angle between the velocity vector and acceleration vector, we can use the dot product formula: $$\cos(\theta) = \frac{\mathbf{v} \cdot \mathbf{a}}{|\mathbf{v}| \cdot |\mathbf{a}|}$$ First, find the dot product of v and a at the specified time: $$\mathbf{v}(t) = \left( 2b\omega\cos\left(\frac{\pi}{2}\right) \mathbf{i}, - b\omega\sin\left(\frac{\pi}{2}\right) \mathbf{j} \right) = (0, -b\omega)$$ $$\mathbf{a}(t) = \left( -2b\omega^2\sin\left(\frac{\pi}{2}\right) \mathbf{i}, - b\omega^2\cos\left(\frac{\pi}{2}\right) \mathbf{j} \right) = (-2b\omega^2, 0)$$ Now find the dot product: $$\mathbf{v} \cdot \mathbf{a} = (0)(-2b\omega^2) + (-b\omega)(0) = 0$$ Next, find the magnitudes of the vectors at the specified time: $$|\mathbf{v}| = \omega b \sqrt{4\cos^2\left(\frac{\pi}{2}\right) + \sin^2\left(\frac{\pi}{2}\right)} = \omega b$$ $$|\mathbf{a}| = \sqrt{(-2b\omega^2\sin(\frac{\pi}{2}))^2 + (- b\omega^2\cos(\frac{\pi}{2}))^2}$$ $$|\mathbf{a}| = \sqrt{4b^2\omega^4\sin^2(\frac{\pi}{2})} = 2b\omega^2$$ Now, calculate the angle: $$\cos(\theta) = \frac{0}{(\omega b)(2b\omega^2)} = 0$$ Since the cosine of 90 degrees is 0, the angle between the velocity and acceleration vectors at \( t = \frac{\pi}{2\omega} \) is 90 degrees or \( \pi / 2 \) radians.

Key Concepts

Velocity Vector

In physics, the velocity vector is an essential component that describes how fast and in which direction a particle is moving. For particles in motion, this vector arises as the first derivative of the position vector with respect to time.

The velocity vector \( \mathbf{v} \) for our elliptical orbit is calculated by differentiating the given position vector \(\mathbf{r} = 2b\sin(\omega t) \mathbf{i} + b\cos(\omega t) \mathbf{j} \). Applying the chain rule, we derive:
\( \mathbf{v} = 2b\omega\cos(\omega t) \mathbf{i} - b\omega\sin(\omega t) \mathbf{j} \).

This vector gives a precise mathematical representation of the particle's change in position over time. Understanding this vector helps us not only to determine the speed of a particle but also how its directional movement varies within the elliptical orbit. A helpful tip is to think of the velocity vector as the arrow that points in the direction the particle is currently heading.

Acceleration Vector

Acceleration vector describes the rate of change of the velocity vector over time. It tells us how the particle's speed and direction change within its elliptical path.

To find the acceleration vector \( \mathbf{a} \), we need to differentiate the velocity vector again with respect to time. Here's how we step through this process: \( \mathbf{a} = \frac{d^2\mathbf{r}}{dt^2} \) results in \( \mathbf{a} = -2b\omega^2\sin(\omega t) \mathbf{i} - b\omega^2\cos(\omega t) \mathbf{j} \).

The negative signs in the components of the acceleration vector indicate opposing motion to the direction of those components of the velocity vector. This suggests that within an elliptical orbit, particles experience changing acceleration due to varying gravitational or centripetal forces. The vector essentially highlights how swiftly a particle can alter its current trajectory.

Particle Speed

Particle speed is a scalar quantity that indicates how fast the particle is moving regardless of its direction. It is the magnitude of the velocity vector.

For the elliptical orbit described, the particle speed is calculated as follows: \( speed = |\mathbf{v}| = \sqrt{4b^2\omega^2\cos^2(\omega t) + b^2\omega^2\sin^2(\omega t)} \).
By factoring out common elements, we simplify this to:\( speed = \omega b \sqrt{4\cos^2(\omega t) + \sin^2(\omega t)} \).

This equation shows that the particle's speed involves both cosine and sine functions, reflecting the elliptical nature of its path. These trigonometric functions contribute to the speed's periodic variation as the particle moves through different points in its orbit.

Dot Product

When determining the angle between the velocity vector and the acceleration vector, the dot product becomes a pivotal tool. The dot product \( \mathbf{v} \cdot \mathbf{a} \) relates two vectors and helps us to find the cosine of the angle between them.

In our problem, the dot product at time \( t = \frac{\pi}{2\omega} \) was calculated as: \( \mathbf{v} \cdot \mathbf{a} = 0 \). The cosine of the angle is then given by:\( \cos(\theta) = \frac{\mathbf{v} \cdot \mathbf{a}}{ |\mathbf{v}| \cdot |\mathbf{a}|} \), simplifying to zero.
This indicates an angle of 90 degrees, illustrating that the velocity and acceleration vectors are perpendicular at this point in the orbit. Situations where these vectors are either perpendicular or parallel can reveal a lot about the motion properties, like the nature of forces acting on the particle.