Question

Consider an object moving according to the position function \(\mathbf{r}(t)=a \cos \omega t \mathbf{i}+a \sin \omega t \mathbf{j}\) Determine the speed of the object at any time \(t\) and explain its value relative to the value of \(a_{\mathrm{T}}\)

Short answer

The speed of the object at any time \(t\) is found to be \(a\omega\), which is constant, and therefore, the tangential acceleration (\(a_{\mathrm{T}}\)) is zero.

Step-by-step solution

Derive the velocity

The velocity \(v(t)\) is the derivative of the position function:\[v(t) = \frac{d \mathbf{r}(t)}{dt} = -a\omega \sin(\omega t) \mathbf{i} + a\omega \cos(\omega t) \mathbf{j}\]

Find the speed

The speed is the magnitude of the velocity vector:\[|\mathbf{v}(t)| = \sqrt{(-a\omega \sin \omega t)^2+(a\omega \cos \omega t)^2}=a\omega\]

Relate the speed with tangential acceleration

Since the speed is constant, this implies the tangetial acceleration \(a_{\mathrm{T}}\) is zero, as there is no change in the speed. Therefore, the value of the speed relative to \(a_{\mathrm{T}}\) is that it imples the \(a_{\mathrm{T}}\) is zero.

Key concepts

Position Function

The position function, denoted as \( \mathbf{r}(t) \), describes the location of an object at any given time \( t \). This powerful tool in calculus is essentially a vector function that maps a specific time to a point in space. In our example, the function is given by \( \mathbf{r}(t)=a \cos \omega t \mathbf{i}+a \sin \omega t \mathbf{j} \).

This equation represents the circular motion of an object on a plane, where \( a \) is the radius of the circle, and \( \omega \) signifies the angular frequency. The components \( \cos \omega t \) and \( \sin \omega t \) are periodic functions that repeat every \( 2\pi \) radians, or one full circle. Consequently, the object's motion described by this function is also periodic, meaning it repeats its path in a cyclical fashion.

Understanding the position function allows us to visualize the object’s pathway. Furthermore, differentiating this function with respect to time gives us the velocity vector, offering insights into the object's instantaneous speed and direction.

Velocity Vector

The velocity vector is a fundamental concept in calculus that extends beyond just a simple scalar value of speed—it encapsulates both magnitude and direction. We form the velocity vector, \( \mathbf{v}(t) \), by differentiating the position function with respect to time. Using our example, we calculate the derivative of \( \mathbf{r}(t) = a \cos \omega t \mathbf{i}+a \sin \omega t \mathbf{j} \), obtaining \[\mathbf{v}(t) = -a\omega \sin(\omega t) \mathbf{i} + a\omega \cos(\omega t) \mathbf{j}\]

The negative sine component and the positive cosine component indicate a change of movement in the opposite direction along the axes, which is characteristic of rotational motion. Delving deeper, the magnitude of the velocity, or the speed of the object, can be found by taking the square root of the sum of the squares of its components, which in our case, simplifies to a constant \( a\omega \). This reveals a crucial property: the speed of an object in circular motion is constant, not affected by the periodic nature of the trigonometric functions from which it is derived.

Velocity vectors are instrumental in not just describing how fast an object is moving, but also in which direction it is travelling at any instant. This information is indispensable for complete comprehension of the object's motion.

Tangential Acceleration

Tangential acceleration, denoted as \( a_{\mathrm{T}} \), is the rate at which an object's velocity changes in magnitude along its path, often associated with changes in the speed of an object moving along a curve. Normally, it's found by taking the derivative of the velocity vector with respect to time. However, in the example where the object's speed is constant, the tangential acceleration is zero. This is because \( a_{\mathrm{T}} \) quantifies the change in speed, and if velocity is constant, then there is no change; hence, no tangential acceleration.

\[a_{\mathrm{T}} = \frac{d|\mathbf{v}(t)|}{dt} = 0\]
The implication of having a zero tangential acceleration reinforces the idea that the object moves at a uniform speed. This constancy is characteristic of uniform circular motion, where despite an object following a curved path, the actual speed along that path doesn’t increase or decrease. In our case, the object’s speed is not relative to its tangential acceleration simply because there is none; \( a_{\mathrm{T}} \) is nonexistent. This aspect of motion is essential, especially in cases of circular motion, where despite the directional change, the speed could remain unaffected.