Problem 11 - 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 \(t\). $$ \mathbf{r}(t)=2 t \mathbf{i}+t^{2} \mathbf{j} $$

Step-by-Step Solution

Answer

Tangential component: \(a_T = \frac{4t}{\sqrt{4 + 4t^2}}\), Normal component: \(a_N = \sqrt{4 - \frac{16t^2}{4 + 4t^2}}\).

Step 1: Find the Velocity Vector

Start by differentiating the position vector \( \mathbf{r}(t) = 2t \mathbf{i} + t^2 \mathbf{j} \) with respect to time \( t \) to get the velocity vector \( \mathbf{v}(t) \). \[\mathbf{v}(t) = \frac{d}{dt}(2t \mathbf{i} + t^2 \mathbf{j}) = 2 \mathbf{i} + 2t \mathbf{j}\]

Step 2: Find the Acceleration Vector

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

Step 3: Compute the Magnitude of the Velocity

Find the magnitude of \( \mathbf{v}(t) = 2 \mathbf{i} + 2t \mathbf{j} \) as it is needed for further calculations.\[|\mathbf{v}(t)| = \sqrt{(2)^2 + (2t)^2} = \sqrt{4 + 4t^2}\]

Step 4: Calculate Tangential Component of Acceleration

The tangential component \( a_T \) of acceleration is given by the derivative of the magnitude of velocity with respect to time.First, find \( \frac{d}{dt} |\mathbf{v}(t)| \):\[|\mathbf{v}(t)| = \sqrt{4 + 4t^2}\]Then,\[\frac{d}{dt} |\mathbf{v}(t)| = \frac{1}{2\sqrt{4 + 4t^2}} \cdot (8t) = \frac{4t}{\sqrt{4 + 4t^2}}\]Hence,\[a_T = \frac{4t}{\sqrt{4 + 4t^2}}\]

Step 5: Calculate Normal Component of Acceleration

The normal component \( a_N \) of acceleration is found using \( a_N = \sqrt{|\mathbf{a}(t)|^2 - a_T^2} \).First, find \(|\mathbf{a}(t)| \):\[|\mathbf{a}(t)| = |2 \mathbf{j}| = 2\]Then,\[a_N = \sqrt{2^2 - \left(\frac{4t}{\sqrt{4 + 4t^2}}\right)^2}\]Simplify:\[a_N = \sqrt{4 - \frac{16t^2}{4 + 4t^2}}\]

Key Concepts

Tangential Acceleration

Tangential acceleration refers to the component of acceleration along the direction of the velocity vector at that instant. Essentially, it measures how fast the speed of a particle is changing as it moves along its path.
To find tangential acceleration, we start by considering the magnitude of the velocity vector. For the position vector \( \mathbf{r}(t) = 2t \mathbf{i} + t^2 \mathbf{j} \), the velocity vector is \( \mathbf{v}(t) = 2 \mathbf{i} + 2t \mathbf{j} \).
Calculate the magnitude of this velocity vector, which is \( |\mathbf{v}(t)| = \sqrt{4 + 4t^2} \). By differentiating this magnitude with respect to time, we obtain the tangential acceleration paralleled to the path:
\[ a_T = \frac{4t}{\sqrt{4 + 4t^2}} \]
This calculation evaluates how the rate of speed of the particle changes over time, detailing the accumulation or reduction of velocity directly along the tangent to the path.

Normal Acceleration

Normal acceleration represents the component of acceleration perpendicular to the velocity vector of the moving particle. It indicates how rapidly the direction of the velocity vector is changing - essentially, a measure of the curvature of the path.
After calculating the tangential component \(a_T\), we use the formula \( a_N = \sqrt{ |\mathbf{a}(t)|^2 - a_T^2 } \) to find the normal acceleration.
For \( \mathbf{a}(t) = 2 \mathbf{j} \), its magnitude is \( |\mathbf{a}(t)| = 2 \). The normal acceleration works out to be:
\[ a_N = \sqrt{4 - \frac{16t^2}{4 + 4t^2}} \]
This provides us with how the path alters its direction as the particle moves, responsible for the turning motion distinct from mere speed changes.

Velocity Vector

The velocity vector of a particle describes how its position changes with time, capturing both the speed and direction.
Starting with the position vector \(\mathbf{r}(t) = 2t \mathbf{i} + t^2 \mathbf{j}\), differentiate it with respect to time to get the velocity vector \(\mathbf{v}(t)\):
\[ \mathbf{v}(t) = \frac{d}{dt}(2t \mathbf{i} + t^2 \mathbf{j}) = 2 \mathbf{i} + 2t \mathbf{j} \]
Here, the components \(2 \mathbf{i} \) and \(2t \mathbf{j} \) are vector projections along the \(x\) and \(y\) axes, respectively. The calculation implies a constant velocity along \(x\) and increasing velocity along \(y\), reflecting motion in the plane that varies over time.

Acceleration Vector

The acceleration vector is derived from the velocity vector and reveals how the velocity of the particle changes over time, influencing both speed and direction.
To find this, differentiate the velocity vector with respect to \(t\):
\[ \mathbf{a}(t) = \frac{d}{dt}(2 \mathbf{i} + 2t \mathbf{j}) = 2 \mathbf{j} \]
The result indicates that only the \(y\) component changes, meaning the particle experiences a constant acceleration perpendicular to the \(x\) axis. This constant value suggests a steady force is acting on the particle, altering its course or speed.