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Consider a particle \end{tabular} moving on a circular path of radius \(b\) described by \(\mathbf{r}(t)=b \cos \omega t \mathbf{i}+b \sin \omega t \mathbf{j}\) where \(\omega=d \theta / d t\) is the constant angular velocity. Find the velocity vector and show that it is orthogonal to \(\mathbf{r}(t)\).

Short Answer

Expert verified
The velocity vector of the particle is given by \(\mathbf{v}(t) = -b \omega \sin \omega t \mathbf{i} + b \omega \cos \omega t \mathbf{j}\). The fact that the dot product of \(\mathbf{v}(t)\) and \(\mathbf{r}(t)\) is zero proves that the velocity vector is indeed orthogonal to the position vector.

Step by step solution

01

Compute the derivative of the position vector

First, we need to compute the derivative of the position vector to find the velocity vector. The velocity vector is given by \(\mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt}\). Differentiating \(\mathbf{r}(t) = b \cos \omega t \mathbf{i} + b \sin \omega t \mathbf{j}\) with respect to \(t\), we get \(\mathbf{v}(t) = -b \omega \sin \omega t \mathbf{i} + b \omega \cos \omega t \mathbf{j}\).
02

Compute the dot product

After we've computed the velocity vector, we need to prove that this vector is orthogonal to the position vector. This can be done by showing that the dot product of these two vectors is zero. The dot product of \(\mathbf{r}(t)\) and \(\mathbf{v}(t)\) is \(\mathbf{r}(t) \cdot \mathbf{v}(t) = (b \cos \omega t \mathbf{i} + b \sin \omega t \mathbf{j}) \cdot (-b \omega \sin \omega t \mathbf{i} + b \omega \cos \omega t \mathbf{j})\). When computed, the result is 0, proving the vectors are orthogonal.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Orthogonality of Vectors
Orthogonality is a fundamental concept in linear algebra where two vectors are considered orthogonal if their dot product equals zero. In simpler terms, it means the vectors are perpendicular to each other.
In the context of circular motion, the position vector \( \mathbf{r}(t) \) describes the location of a particle on a circular path, while the velocity vector \( \mathbf{v}(t) \) describes how fast and in what direction the particle is moving.
To determine orthogonality, we calculate the dot product:
  • The position vector \( \mathbf{r}(t) = b \cos \omega t \mathbf{i} + b \sin \omega t \mathbf{j} \)
  • The velocity vector \( \mathbf{v}(t) = -b \omega \sin \omega t \mathbf{i} + b \omega \cos \omega t \mathbf{j} \)
After computing the dot product, we find it is indeed zero, proving orthogonality. The vectors' orthogonality indicates that the velocity (or motion) is always tangent to the path of the particle, a distinctive feature of circular motion.
Position Vector
The position vector \( \mathbf{r}(t) \) helps us to determine the location of a particle in space as a function of time. It’s a dynamic expression using coordinates to express movement:
  • For a particle moving in a circle of radius \( b \), the position is given by \( \mathbf{r}(t) = b \cos \omega t \mathbf{i} + b \sin \omega t \mathbf{j} \)
Each component of the position vector represents the projection of the radius on the respective axes.
The term \( b \) specifies the constant radius, indicating how far from the origin the particle is and \( \omega \) relates to the angular velocity which determines how the particle's angle changes over time. The vector \( \mathbf{r}(t) \), therefore, maps out a circle path, demonstrating the motion graphically as time advances.
Understanding the position vector is crucial for grasping more intricate ideas, such as velocity and acceleration in circular motion.
Angular Velocity
Angular velocity \( \omega \) is a fundamental concept describing how fast an object rotates or revolves relative to another point. It is typically expressed in radians per second and captures the rate of change of the angle with respect to time:
  • The formula \( \omega = \frac{d \theta}{dt} \) represents this rate.
For a particle moving in a circle, angular velocity affects both how quickly it traverses its path and the direction of the velocity vector.
In the position vector \( \mathbf{r}(t) = b \cos \omega t \mathbf{i} + b \sin \omega t \mathbf{j} \), \( \omega \) modulates the sine and cosine functions, introducing periodic motion typical of a circular path.
Importantly, it dictates how the velocity vector changes over time, allowing the particle to move smoothly along the circle's tangent without speeding up or slowing down.

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