Problem 44 - Calculus III

Question

The force on a particle is given by \(\mathbf{f}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}\). The particle is located at point \((c, 0)\) at \(t=0\). The initial velocity of the particle is given by \(\mathbf{v}(0)=v_{0} \mathbf{j}\). Find the path of the particle of mass \(\mathbf{m} \cdot(\operatorname{Recall}, \mathbf{F}=m \cdot \mathbf{a}\).)

Step-by-Step Solution

Answer

The particle follows a path defined by the function \( \mathbf{r}(t) = (c+\frac{1}{m} (1-\cos t)) \mathbf{i} + (\frac{1}{m} t - \sin t) \mathbf{j} \).

Step 1: Express Force and Acceleration

The force acting on the particle is given by \( \mathbf{f}(t) = (\cos t) \mathbf{i} + (\sin t) \mathbf{j} \). According to Newton's second law, the force \( \mathbf{F} \) is equal to mass \( m \) times acceleration \( \mathbf{a} \). Thus, we write \( \mathbf{f}(t) = m \cdot \mathbf{a}(t) \), which means \[ \mathbf{a}(t) = \frac{1}{m}[(\cos t) \mathbf{i} + (\sin t) \mathbf{j}] \].

Step 2: Integrate Acceleration to Find Velocity

To find the velocity \( \mathbf{v}(t) \), integrate \( \mathbf{a}(t) \) with respect to \( t \): \[ \mathbf{v}(t) = \int \mathbf{a}(t) \, dt = \frac{1}{m} \left( \int (\cos t) \, dt \right) \mathbf{i} + \frac{1}{m} \left( \int (\sin t) \, dt \right) \mathbf{j} \]. Solving these integrals: \[ \mathbf{v}(t) = \frac{1}{m} (\sin t) \mathbf{i} - \frac{1}{m} (\cos t) \mathbf{j} + \mathbf{C} \], where \( \mathbf{C} \) is the constant of integration.

Step 3: Apply Initial Velocity Condition

Given the initial velocity \( \mathbf{v}(0) = v_0 \mathbf{j} \), substitute \( t = 0 \) into the velocity equation:\[ \mathbf{v}(0) = \frac{1}{m} (\sin 0) \mathbf{i} - \frac{1}{m} (\cos 0) \mathbf{j} + \mathbf{C} = v_0 \mathbf{j} \].Simplifying gives \( \mathbf{C} = \left(0 \right) \mathbf{i} + \left(\frac{1}{m} - v_0 \right) \mathbf{j} \).Thus, \[ \mathbf{v}(t) = \frac{1}{m} (\sin t) \mathbf{i} - \frac{1}{m} (\cos t) \mathbf{j} + \frac{1}{m} \mathbf{j} - v_0 \mathbf{j} \].

Step 4: Integrate Velocity to Find Position

Integrate \( \mathbf{v}(t) \) to find the position \( \mathbf{r}(t) \):\[ \mathbf{r}(t) = \int \mathbf{v}(t) \, dt = \int \left( \frac{1}{m} \sin t \right) \mathbf{i} - \left( \frac{1}{m} \cos t \right) \mathbf{j} + \mathbf{k} \mathbf{j} \, dt \], where \( \mathbf{k} = \frac{1}{m} - v_0 \).Solving gives:\[ \mathbf{r}(t) = -\frac{1}{m} \cos t \mathbf{i} - \frac{1}{m} \sin t \mathbf{j} + \mathbf{k} t \mathbf{j} + \mathbf{D} \], where \( \mathbf{D} \) is the new constant of integration.

Step 5: Apply Initial Position Condition

The particle's initial position is \((c,0)\). At \(t=0\), \[ \mathbf{r}(0) = -\frac{1}{m} \cos 0 \mathbf{i} - \frac{1}{m} \sin 0 \mathbf{j} + \mathbf{D} = c \mathbf{i} \].Thus \( \mathbf{D} = \left(c + \frac{1}{m} \right) \mathbf{i} \), simplifying to\[ \mathbf{r}(t) = \left(c + \frac{1}{m} - \frac{1}{m} \cos t \right) \mathbf{i} \mathbf{k} t \mathbf{j} - \frac{1}{m} \sin t \mathbf{j} \].

Key Concepts

Force and Acceleration

In physics, force is directly linked to acceleration by Newton's second law of motion. This law tells us that the force applied to an object is equal to its mass multiplied by the acceleration it undergoes. The formula is expressed as \( \mathbf{F} = m \cdot \mathbf{a} \). This simple but powerful idea is key to understanding particle motion.

For a particle with a mass \( m \), the force \( \mathbf{f}(t) \) acting on it over time \( t \) can be described by separate components in different directions, such as \( \mathbf{i} \) and \( \mathbf{j} \) in 2D space. In our given problem, the force is \( (\cos t) \mathbf{i} + (\sin t) \mathbf{j} \). Now, by dividing this force by the particle's mass \( m \), we get the acceleration as:

• \( \mathbf{a}(t) = \frac{1}{m}[(\cos t) \mathbf{i} + (\sin t) \mathbf{j}] \)

This formula tells us how the particle's velocity changes over time due to the applied forces. Each component of the acceleration affects the velocity in its direction.

Velocity Integration

After finding acceleration, the next step is to determine how this affects the velocity of the particle. This is achieved by integrating the acceleration function. The integration process basically accumulates all the small changes in speed over time, leading to the velocity function. The integration can be seen in the formula:

• \( \mathbf{v}(t) = \int \mathbf{a}(t) \, dt = \frac{1}{m} \left( \int (\cos t) \, dt \right) \mathbf{i} + \frac{1}{m} \left( \int (\sin t) \, dt \right) \mathbf{j} \)

Solving these integrals involves finding the antiderivatives of the sine and cosine functions. This means:

• \( \mathbf{v}(t) = \frac{1}{m} (\sin t) \mathbf{i} - \frac{1}{m} (\cos t) \mathbf{j} + \mathbf{C} \)

Here, \( \mathbf{C} \) is a constant of integration, which needs to be determined using initial conditions. This result describes how the velocity changes, taking into account any predefined starting values.

Initial Conditions

Initial conditions are crucial as they allow us to determine the constants of integration that appear after solving differential equations. In our scenario, we have initial conditions for both velocity and position which need to be applied.

Initially, the velocity \( \mathbf{v}(0) \) is specified as \( v_0 \mathbf{j} \), meaning there's an initial speed in the \( \mathbf{j} \) direction but not in the \( \mathbf{i} \) direction. Integrating acceleration gives a velocity that includes \( \mathbf{C} \), the constant vector, which we adjust using:

• \( \mathbf{v}(0) = v_0 \mathbf{j} = 0 \mathbf{i} - \frac{1}{m} \mathbf{j} + \mathbf{C} \)

Thus, \( \mathbf{C} \) becomes \( 0 \mathbf{i} + \left( \frac{1}{m} + v_0 \right) \mathbf{j} \).

Similarly, for position, we have an initial point \( (c, 0) \) at \( t=0 \). Just like with velocity, integrate the velocity function to find position, adjusting with a new constant \( \mathbf{D} \), set as follows:

• \( \mathbf{r}(0) = c \mathbf{i} = -\frac{1}{m} \mathbf{i} + \mathbf{D} \)

The position function emerges as a combination of the motion due to integrated velocity and the initial location.