Problem 60 - Fundamentals of Physics

Question

A centripetal-acceleration addict rides in uniform circular motion with radius \(r=3.00 \mathrm{~m}\). At one instant his acceleration is \(\vec{a}=\left(6.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(-4.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}}\). At that instant, what are the val- ues of (a) \(\vec{v} \cdot \vec{a}\) and (b) \(\vec{r} \times \vec{a}\) ?

Step-by-Step Solution

Answer

(a) 0, (b) \(-(18 \sin \theta + 12 \cos \theta)\)

Step 1: Identify the required values for velocity

In uniform circular motion, the velocity vector \( \vec{v} \) is always perpendicular to the acceleration vector \( \vec{a} \). Therefore, the dot product \( \vec{v} \cdot \vec{a} = 0 \).

Step 2: Calculate the position vector

Given the radius \( r = 3.00 \text{ m} \) and an acceleration \( \vec{a} = (6.00 \text{ m/s}^2) \hat{i} + (-4.00 \text{ m/s}^2) \hat{j} \), the position vector \( \vec{r} \) is \( 3.00 \text{ m} \left( \cos \theta \hat{i} + \sin \theta \hat{j} \right) \).

Step 3: Calculate the cross product

The cross product \( \vec{r} \times \vec{a} \) can be calculated as:\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \3 \cos \theta & 3 \sin \theta & 0 \6 & -4 & 0 \end{vmatrix}\]Solving the determinant, we get \( \vec{r} \times \vec{a} = \left(0\hat{i} + 0\hat{j} + (-18 \sin \theta - 12 \cos \theta)\hat{k}\right) \).To find the magnitude, it results in:\[R \cdot A = 18 \intercal \theta + 12\intercal \theta \]

Step 4: Conclusion

From step 3, the value of \( \vec{v} \cdot \vec{a} = 0 \), and \( \vec{r} \times \vec{a}= \left[-(18 \sin \theta + 12 \cos \theta) \right] \).

Key Concepts

uniform circular motion

Uniform circular motion is a type of motion where an object moves along a circular path at a constant speed. This might sound a bit tricky, but when you break it down, it's quite simple! Even though the speed remains constant, the direction of the object's velocity is continuously changing as it moves around the circle. This constant change in direction means there is an acceleration present, known as centripetal acceleration.
In the case of our problem, the centripetal acceleration is what keeps the centripetal-acceleration addict moving in a circle. It acts towards the center of the circle, maintaining the motion's uniformity. The radius of the circle is given as 3.00 meters, which is the constant distance from the center of the circle to the path of motion.
Understanding uniform circular motion is critical when analyzing problems involving centripetal force and acceleration, as these are necessary to keep an object moving in a circular path without changing speed.

velocity vector

The velocity vector, denoted as \( \vec{v} \), is an important concept in describing motion. This vector doesn't just give the speed of an object, but also the direction of that speed. In the context of uniform circular motion, the velocity vector is always tangent to the circle, meaning it's directed along the path of the circle at any given point.
This perpendicular relationship between the velocity and acceleration vectors in circular motion is why their dot product \( \vec{v} \cdot \vec{a} \) is zero. The dot product measures how much one vector goes in the direction of another, and since they're perpendicular, this value is naturally zero. So, when we say \( \vec{v} \cdot \vec{a} = 0 \), it confirms the perpendicularity of the two vectors, a crucial part of circular motion mechanics.
Understanding how the velocity vector behaves and interacts with other vectors like acceleration helps us grasp how objects move in circular paths.

cross product

The cross product, represented by \( \vec{r} \times \vec{a} \), is another vector operation but results in a vector perpendicular to the plane defined by \( \vec{r} \) and \( \vec{a} \). This operation is used to determine the torque or rotational effect of forces.
In the case of our exercise, we use the cross product to find how the position vector \( \vec{r} \) and the acceleration vector \( \vec{a} \) interact in a rotational frame. The vector product produces a new vector along the direction perpendicular to both \( \vec{r} \) and \( \vec{a} \).
The cross product's magnitude is given by the area of the parallelogram spanned by the two vectors. It reflects the extent to which the acceleration can cause rotational motion about the center of the circle. Here, after doing the math, we find \( \vec{r} \times \vec{a} = -(18 \sin \theta + 12 \cos \theta) \hat{k} \) indicating the rotational influence at any angle \( \theta \).

dot product

The dot product, unlike the cross product, results in a scalar quantity and is determined by the product of the magnitudes of two vectors and the cosine of the angle between them. In the simple calculation of \( \vec{v} \cdot \vec{a} \), where \( \vec{v} \) and \( \vec{a} \) are velocity and acceleration vectors respectively, the resultant is zero when the vectors are perpendicular.
In uniform circular motion, this perpendicular nature is key because it ensures energy conservation throughout the motion. The dot product's result being zero signifies no work done along the direction of motion, maintaining constant kinetic energy of the moving object. Furthermore, it highlights that any force acting in the direction of velocity doesn't alter speed, only direction.
This fundamental property of the dot product ensures the mathematical consistency in problems involving circular motion and helps in simplifying complex dynamic equations.