Problem 13 - Advanced Engineering Mathematics

Question

In Problems \(7-16, \mathbf{r}(t)\) is the position vector of a moving particle. Find the tangential and normal components of the acceleration at any \(t\). $$ \mathbf{r}(t)=2 \cos t 1+2 \sin t \mathrm{~J} $$

Step-by-Step Solution

Answer

The tangential component of acceleration is 0, and the normal component is 2.

Step 1: Find the velocity vector

First, take the derivative of the position vector, \( \mathbf{r}(t) = 2 \cos t \, \mathbf{i} + 2 \sin t \, \mathbf{j} \), with respect to \( t \). The derivative \( \mathbf{v}(t) \) represents the velocity vector.\[ \mathbf{v}(t) = \frac{d}{dt}(2 \cos t \, \mathbf{i} + 2 \sin t \, \mathbf{j}) = -2 \sin t \, \mathbf{i} + 2 \cos t \, \mathbf{j} \]

Step 2: Find the acceleration vector

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

Step 3: Find the speed of the particle

Calculate the magnitude of the velocity vector \( \mathbf{v}(t) \) to obtain the speed \( v(t) \).\[ v(t) = \| \mathbf{v}(t) \| = \sqrt{(-2 \sin t)^2 + (2 \cos t)^2} = \sqrt{4 \sin^2 t + 4 \cos^2 t} = \sqrt{4} = 2 \]

Step 4: Find the tangential component of acceleration

The tangential component of the acceleration can be found using the formula \( a_T = \frac{d}{dt}(v(t)) \). Since the speed \( v(t) = 2 \) is constant, its derivative with respect to \( t \) is zero.\[ a_T = \frac{d}{dt}(2) = 0 \]

Step 5: Find the normal component of acceleration

The normal component of acceleration is given by \( a_N = \sqrt{a^2 - a_T^2} \), where \( a = \| \mathbf{a}(t) \| \). Since \( a_T = 0 \), \( a_N = a \).First, calculate \( \| \mathbf{a}(t) \| \):\[ \| \mathbf{a}(t) \| = \sqrt{(-2 \cos t)^2 + (-2 \sin t)^2} = \sqrt{4 \cos^2 t + 4 \sin^2 t} = \sqrt{4} = 2 \]Thus, \( a_N = 2 \).

Key Concepts

Tangential Acceleration

Tangential acceleration refers to the component of an object's acceleration that is parallel to the direction of its velocity vector. In simpler terms, it's the part of the acceleration that changes the speed of the object along the path it's traveling.
For this particular problem, we find that the speed or magnitude of velocity, denoted as \( v(t) \), is constant at 2. This indicates that the object moves around a circular path with constant speed. Therefore, the tangential acceleration is the rate of change of this speed. Since the speed doesn't change over time (it's constant), the tangential acceleration \( a_T \) is zero.
In vector calculus, the formula used for tangential acceleration is \( a_T = \frac{d}{dt}(v(t)) \). For any particle with constant speed, no matter the path, this derivative is zero, leading to a tangential acceleration of zero.

Normal Acceleration

Normal acceleration, sometimes called centripetal acceleration, is the component of acceleration that is perpendicular to the velocity vector. This part of acceleration changes the direction of the velocity vector and keeps the object moving along a curved path.
As shown in the solution, the normal acceleration \( a_N \) can be calculated using the formula \( a_N = \sqrt{a^2 - a_T^2} \). In this problem, because the tangential acceleration \( a_T \) is zero, normal acceleration \( a_N \) is simply the magnitude of the total acceleration \( a \). We find \( a_N = 2 \), which confirms that all of the particle's acceleration is directed towards changing its direction on the circular path.

• This emphasizes that normal acceleration is responsible for maintaining circular motion.
• Even with zero tangential acceleration, objects can still experience acceleration through turning or changing direction.

Velocity Vector

The velocity vector represents both the speed and the direction of a moving object at any given time. For this exercise, it's crucial to first determine this vector in order to explore other aspects of motion.
Calculated as the derivative of the position vector \( \mathbf{r}(t) \), the velocity vector for this particle is \( \mathbf{v}(t) = -2 \sin t \, \mathbf{i} + 2 \cos t \, \mathbf{j} \). This shows that the particle moves with components in both the \( \mathbf{i} \) (x-direction) and \( \mathbf{j} \) (y-direction).
The length or magnitude of this vector gives us the speed, which remains constant at 2, as seen in the solution. This constancy reflects uniform circular motion around a fixed path at a fixed speed.

Acceleration Vector

The acceleration vector is derived by differentiating the velocity vector with respect to time. It provides insight into how the velocity vector is changing, revealing details about the speed and direction of motion.
Here, differentiating the velocity vector \( \mathbf{v}(t) \) results in the acceleration vector \( \mathbf{a}(t) = -2 \cos t \, \mathbf{i} - 2 \sin t \, \mathbf{j} \). This tells us about the changes in motion occurring at each point on the trajectory.
The acceleration vector shows us the combination of tangential and normal accelerations: in this case, the acceleration is entirely normal due to constant speed (tangential acceleration is zero), and its magnitude \( \| \mathbf{a}(t) \| \) is 2.

• An acceleration vector not only reveals the type of motion but also complements our understanding of forces acting on the particle in motion.
• In uniform circular motion, acceleration is always directed perpendicular to the velocity vector.