Problem 8 - Advanced Engineering Mathematics

Question

\(\mathbf{r}(t)\) is the position vector of a moving particle. Find the tangential and normal components of the acceleration at any \(t\). $$ \mathbf{r}(t)=3 \cos t \mathbf{i}+2 \sin t \mathbf{j}+t \mathbf{k} $$

Step-by-Step Solution

Answer

The tangential component is \( a_T = \frac{d}{dt}(|\mathbf{v}(t)|) \) and the normal component is \( a_N = \sqrt{|\mathbf{a}(t)|^2 - a_T^2} \).

Step 1: Find the velocity vector

Start by differentiating the position vector \( \mathbf{r}(t) \) with respect to \( t \).\[ \mathbf{v}(t) = \frac{d}{dt}[3 \cos t \mathbf{i} + 2 \sin t \mathbf{j} + t \mathbf{k}] = -3 \sin t \mathbf{i} + 2 \cos t \mathbf{j} + \mathbf{k} \]

Step 2: Find the acceleration vector

Differentiate the velocity vector \( \mathbf{v}(t) \) with respect to \( t \) to get the acceleration vector.\[ \mathbf{a}(t) = \frac{d}{dt}[-3 \sin t \mathbf{i} + 2 \cos t \mathbf{j} + \mathbf{k}] = -3 \cos t \mathbf{i} - 2 \sin t \mathbf{j} \]

Step 3: Find the speed

The speed is the magnitude of the velocity vector \( \mathbf{v}(t) \).\[ |\mathbf{v}(t)| = \sqrt{(-3 \sin t)^2 + (2 \cos t)^2 + 1^2} = \sqrt{9 \sin^2 t + 4 \cos^2 t + 1} \]

Step 4: Simplify speed

Simplify the expression for \( |\mathbf{v}(t)| \).\[ |\mathbf{v}(t)| = \sqrt{9 \sin^2 t + 4 \cos^2 t + 1} \approx \sqrt{5 + 5 \sin^2 t} \]

Step 5: Tangential component of acceleration

The tangential component of acceleration \( a_T \) is given by the rate of change of speed with respect to time \( t \).\[ a_T = \frac{d}{dt}(|\mathbf{v}(t)|) \]

Step 6: Normal component of acceleration

The normal component \( a_N \) is given by \( a_N = \sqrt{|\mathbf{a}(t)|^2 - a_T^2} \), where \( |\mathbf{a}(t)| \) is the magnitude of the acceleration.\[ |\mathbf{a}(t)| = \sqrt{(-3 \cos t)^2 + (-2 \sin t)^2} = \sqrt{9 \cos^2 t + 4 \sin^2 t} \]

Key Concepts

Tangential Component

The tangential component of the acceleration, often denoted as \( a_T \), represents how fast the speed of a particle is changing along the path it travels. It is directly linked to the rate of change of the magnitude of the velocity vector. In simple words, it tells us how the particle accelerates or decelerates as it follows its trajectory.
To find \( a_T \), calculate the derivative of the speed \( |\mathbf{v}(t)| \) with respect to time \( t \). This approach focuses purely on changes in speed, leaving out any change in direction.
For example, in the original exercise, the tangential component is found using the expression:

• \( a_T = \frac{d}{dt}(|\mathbf{v}(t)|) \)

Understanding \( a_T \) is valuable when analyzing movements in a straight line or around curves because it is purely a measure of how fast or slow a particle is moving at any point.

Normal Component

The normal component of acceleration, denoted as \( a_N \), captures how quickly the direction of a particle changes as it moves. While the tangential component relates to speed changes, \( a_N \) is associated with the particle's path curvature.
This component reflects how much the acceleration is pushing the particle towards or pulling it away from the path it is on, essentially guiding its turning motion. In mathematical terms, it can be calculated by using the formula:

• \( a_N = \sqrt{|\mathbf{a}(t)|^2 - a_T^2} \)

Where \(|\mathbf{a}(t)|\) is the magnitude of the total acceleration vector.
It is crucial for understanding phenomena where the path changes sharply, like a car turning sharply or a roller coaster loop.

Acceleration Vector

The acceleration vector \( \mathbf{a}(t) \) describes how the velocity of a particle changes over time. It provides a complete picture of how a particle speeds up, slows down, or changes direction.
This vector is found by differentiating the velocity vector \( \mathbf{v}(t) \) with respect to time. It can have components both in the direction of the movement (tangential) and perpendicular to it (normal).
Using the original example:

• Acceleration vector \( \mathbf{a}(t) = \frac{d}{dt}\[-3 \sin t \mathbf{i} + 2 \cos t \mathbf{j} + \mathbf{k}\] = -3 \cos t \mathbf{i} - 2 \sin t \mathbf{j} \).

This means the acceleration vector can point in various directions, dictating both speed and path alterations.

Velocity Vector

The velocity vector \( \mathbf{v}(t) \) of a particle informs about its speed and direction at any given moment. Imagine it as an arrow pointing where the particle is moving, with its length indicating how fast it moves.
To determine it, differentiate the position vector \( \mathbf{r}(t) \) with respect to time. This gives us both the components:

• Velocity vector \( \mathbf{v}(t) = \frac{d}{dt}[3 \cos t \mathbf{i} + 2 \sin t \mathbf{j} + t \mathbf{k}] = -3 \sin t \mathbf{i} + 2 \cos t \mathbf{j} + \mathbf{k} \).

Here, each unit vector \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) indicates movement in the respective coordinate axes.
By having this vector, one can precisely analyze how swift and in which direction a particle advances at any specific instant.