Problem 57 - Physics for Scientists and Engineers with Modern Physics

Question

(III) The position of a particle moving in the \(x y\) plane is given by \(\overrightarrow{\mathbf{r}}=2.0 \cos (3.0 \mathrm{rad} / \mathrm{s} t) \hat{\mathbf{i}}+2.0 \sin (3.0 \mathrm{rad} / \mathrm{s} t) \hat{\mathbf{j}}\) where \(r\) is in meters and \(t\) is in seconds. ( \(a\) ) Show that this represents circular motion of radius \(2.0 \mathrm{~m}\) centered at the origin. (b) Determine the velocity and acceleration vectors as functions of time. ( \(c\) ) Determine the speed and magnitude of the acceleration. \((d)\) Show that \(a=v^{2} / r .(e)\) Show that the acceleration vector always points toward the center of the circle.

Step-by-Step Solution

Answer

The particle moves in a circle of radius 2 m; \(v = 6 \), \(a = 18 \), \(a = v^2/r\), pointing to the center.

Step 1: Verify Circular Motion

To confirm that the motion is circular, we should express the position vector \(\overrightarrow{\mathbf{r}}\) using a parametric form for a circle. The parametric equations for a circle of radius \(R\) are \(x = R \cos(\omega t)\) and \(y = R \sin(\omega t)\). Comparing this with \(\overrightarrow{\mathbf{r}} = 2.0 \cos(3.0 \mathrm{rad}/\mathrm{s} \cdot t) \hat{\mathbf{i}} + 2.0 \sin(3.0 \mathrm{rad}/\mathrm{s} \cdot t) \hat{\mathbf{j}}\), it fits the form with \(R = 2.0\,\mathrm{m}\), confirming it is circular with center \((0,0)\) and radius \(2.0\,\mathrm{m}\).

Step 2: Calculate Velocity Vector

The velocity vector \(\overrightarrow{\mathbf{v}}\) is found by differentiating the position vector \(\overrightarrow{\mathbf{r}}\) with respect to time \(t\). Thus, \(\overrightarrow{\mathbf{v}} = \frac{d}{dt}(2.0 \cos(3.0t)\hat{\mathbf{i}} + 2.0 \sin(3.0t)\hat{\mathbf{j}}) = -6.0 \sin(3.0t)\hat{\mathbf{i}} + 6.0 \cos(3.0t)\hat{\mathbf{j}}\).

Step 3: Calculate Acceleration Vector

The acceleration vector \(\overrightarrow{\mathbf{a}}\) is the derivative of the velocity vector \(\overrightarrow{\mathbf{v}}\) with respect to time. Differentiating \(\overrightarrow{\mathbf{v}} = -6.0 \sin(3.0t)\hat{\mathbf{i}} + 6.0 \cos(3.0t)\hat{\mathbf{j}}\), we get \(\overrightarrow{\mathbf{a}} = -18.0 \cos(3.0t)\hat{\mathbf{i}} -18.0 \sin(3.0t)\hat{\mathbf{j}}\).

Step 4: Determine Speed and Magnitude of Acceleration

The speed \(v\) is the magnitude of the velocity vector: \(v = \sqrt{(-6.0 \sin(3.0t))^2 + (6.0 \cos(3.0t))^2} = 6.0 \). The magnitude of the acceleration is \(a = \sqrt{(-18.0 \cos(3.0t))^2 + (-18.0 \sin(3.0t))^2} = 18.0 \).

Step 5: Show that \(a = \frac{v^2}{r}\)

We know \(v = 6.0\) and \(r = 2.0\,\mathrm{m}\). Substituting into \(a = \frac{v^2}{r}\), we have \(a = \frac{6.0^2}{2.0} = 18.0\), which matches the computed acceleration magnitude.

Step 6: Direction of Acceleration Vector

The acceleration vector \(\overrightarrow{\mathbf{a}} = -18.0 \cos(3.0t)\hat{\mathbf{i}} -18.0 \sin(3.0t)\hat{\mathbf{j}}\) can be rewritten as \(-18.0 \overrightarrow{\mathbf{r}}\). This shows it points in the opposite direction of the position vector, indicating it always points toward the center of the circle.

Key Concepts

Velocity Vector

In circular motion, understanding the velocity vector is crucial to describe how fast and in what direction an object is moving at any given point along its path. The velocity vector is the derivative of the position vector with respect to time. By differentiating the particle's position function, \(\overrightarrow{\mathbf{r}} = 2.0 \cos(3.0t)\hat{\mathbf{i}} + 2.0 \sin(3.0t)\hat{\mathbf{j}}\), we find the expression for the velocity vector:\(\overrightarrow{\mathbf{v}} = -6.0 \sin(3.0t)\hat{\mathbf{i}} + 6.0 \cos(3.0t)\hat{\mathbf{j}}\).

This result indicates that the velocity is perpendicular to the radius. Let's look at some key takeaways from the velocity vector:

• The velocity magnitude remains constant, as expected for uniform circular motion.
• The velocity vector is tangential to the circular path at every instant.
• Its direction changes continuously, keeping the motion circular.

Acceleration Vector

The acceleration vector helps us understand how the velocity of the particle changes over time. Differentiating the velocity vector \(\overrightarrow{\mathbf{v}} = -6.0 \sin(3.0t)\hat{\mathbf{i}} + 6.0 \cos(3.0t)\hat{\mathbf{j}}\), we discover the acceleration vector:\(\overrightarrow{\mathbf{a}} = -18.0 \cos(3.0t)\hat{\mathbf{i}} -18.0 \sin(3.0t)\hat{\mathbf{j}}\).

Let's break down what this means:

• The acceleration vector indicates a change in direction rather than speed, typical of circular motion.
• It points directly toward the center of the circle, demonstrating a centripetal acceleration.
• Its magnitude is constant, calculated as \(18.0\), consistent with uniform circular motion.

Parametric Equations

Parametric equations are a powerful tool in describing the positions of points moving along a path, such as a circle in two-dimensional space. In this exercise, the particle's position is given by two parametric equations:

• \(x(t) = 2.0 \cos(3.0t)\)
• \(y(t) = 2.0 \sin(3.0t)\)

These equations specify that the particle traces out a circle with a radius of 2 meters.

Some benefits of using parametric equations include:

• Providing a clear representation of a motion path beyond simple coordinate descriptions.
• Allowing easy differentiation to find velocity and acceleration vectors.
• Offering great flexibility in modeling different paths by adjusting parameters, like \(\omega = 3.0 \text{ rad/s}\) in our example.

Motion Dynamics

The dynamics of motion in circular paths involve understanding the relationship between different quantities. Here we delve into the link between speed, acceleration magnitude, and radius in circular motion dynamics.

• Speed in circular motion is given by the magnitude of the velocity vector, denoted as \(v = 6.0\) (in this example).
• The magnitude of acceleration is determined by the centripetal force required to maintain circular motion, found to be \(18.0\).

One fundamental dynamic relationship is that the acceleration required for circular motion is proportional to the square of the speed \(v\) and inversely proportional to the radius \(r\):\[ a = \frac{v^2}{r} \].Substituting our values shows this relationship holds true - a crucial insight into the understanding of motion dynamics.

These dynamics explain the continuous change in velocity direction while maintaining constant speed, characteristic of uniform circular motion.