Problem 10 - Classical Mechanics

Question

A particle moves in a circle (center \(O\) and radius \(R\) ) with constant angular velocity \(\omega\) counter-clockwise. The circle lies in the \(x y\) plane and the particle is on the \(x\) axis at time \(t=0 .\) Show that the particle's position is given by $$\mathbf{r}(t)=\hat{\mathbf{x}} R \cos (\omega t)+\hat{\mathbf{y}} R \sin (\omega t)$$ Find the particle's velocity and acceleration. What are the magnitude and direction of the acceleration? Relate your results to well-known properties of uniform circular motion.

Step-by-Step Solution

Answer

The particle's velocity is \( \mathbf{v}(t) = \hat{\mathbf{x}} (-R\omega \sin(\omega t)) + \hat{\mathbf{y}} (R\omega \cos(\omega t)) \), and its acceleration is \( \mathbf{a}(t) = -R \omega^2 (\cos(\omega t) \hat{\mathbf{x}} + \sin(\omega t) \hat{\mathbf{y}}) \), with magnitude \( R\omega^2 \), directed towards the circle's center.

Step 1: Parametrize the Circular Motion

The motion of a particle in a circle can be described using parametric equations. Since at time \(t=0\), the particle is on the \(x\)-axis, we can express its position in the plane at any subsequent time as: \(x(t) = R \cos(\omega t)\) and \(y(t) = R \sin(\omega t)\). These functions use the fact that a circle can be represented in terms of trigonometric functions.

Step 2: Express the Position Vector

Combine the parametric expressions into a single position vector \( \mathbf{r}(t) \). Thus, \( \mathbf{r}(t) = \hat{\mathbf{x}} R \cos(\omega t) + \hat{\mathbf{y}} R \sin(\omega t) \). This expression describes the particle's position at any time \( t \) in terms of its unit vectors and corresponding coordinates.

Step 3: Differentiate to Find Velocity

To find the velocity vector \( \mathbf{v}(t) \), differentiate the position vector \( \mathbf{r}(t) \) with respect to time \( t \). Since \( \frac{d}{dt} (\cos(\omega t)) = -\omega \sin(\omega t) \) and \( \frac{d}{dt} (\sin(\omega t)) = \omega \cos(\omega t) \), differentiating gives: \( \mathbf{v}(t) = \hat{\mathbf{x}} (-R\omega \sin(\omega t)) + \hat{\mathbf{y}} (R\omega \cos(\omega t)) \).

Step 4: Differentiate Velocity to Find Acceleration

Differentiate the velocity vector \( \mathbf{v}(t) \) to find the acceleration vector \( \mathbf{a}(t) \). So, \( \mathbf{a}(t) = \frac{d}{dt}(-R\omega \sin(\omega t)) \hat{\mathbf{x}} + \frac{d}{dt}(R\omega \cos(\omega t)) \hat{\mathbf{y}} \), which results in \( \mathbf{a}(t) = -R \omega^2 \cos(\omega t) \hat{\mathbf{x}} - R \omega^2 \sin(\omega t) \hat{\mathbf{y}} \).

Step 5: Magnitude and Direction of Acceleration

To find the magnitude of the acceleration, calculate \( |\mathbf{a}(t)| = \sqrt{(-R \omega^2 \cos(\omega t))^2 + (-R \omega^2 \sin(\omega t))^2} \). Simplifying, we get \( |\mathbf{a}(t)| = R \omega^2 \). The direction of \( \mathbf{a}(t) \) is opposite to \( \mathbf{r}(t) \), indicating centripetal acceleration toward the center \( O \).

Step 6: Relate to Uniform Circular Motion

In uniform circular motion, the particle remains at a constant speed and experiences centripetal acceleration directed towards the circle's center. Our results show that the magnitude of this acceleration is \( R\omega^2 \), consistent with the general formula for centripetal acceleration \( a_c = \frac{v^2}{R} \). In this scenario, \( v = R\omega \), thus confirming the properties of circular motion.

Key Concepts

Parametric Equations

Parametric equations are a powerful tool for describing the motion of a particle in space, especially when dealing with circular or elliptical paths. In this exercise, the particle's path is circular with a constant radius. To express the position of the particle at any given time using parametric equations, we rely on the trigonometric functions of cosine and sine. These functions provide a natural way to represent circular motion because they describe the periodic nature of the particle's path.

• The position in the x-direction is given by: \(x(t) = R \cos(\omega t)\).
• The position in the y-direction is expressed as: \(y(t) = R \sin(\omega t)\).

These equations ensure that as time progresses, the particle moves along the circumference of the circle. By combining these two parametric equations, the position vector \(\mathbf{r}(t)\) is formed, describing the particle's location in the 2D plane of motion. It's essential to remember that this approach leverages the cyclic properties of sine and cosine to capture the essence of continuous circular motion.

Centripetal Acceleration

In uniform circular motion, a particle moves at a constant speed along a circular path but is still undergoing acceleration, known as centripetal acceleration. This type of acceleration always points towards the center of the circle and is crucial for maintaining the particle's pathway. Even though the speed is constant, the velocity vector changes due to its continually varying direction.
To quantify centripetal acceleration, consider:

• The formula is \( a_c = R \omega^2 \), where \(R\) is the radius, and \(\omega\) is the angular velocity.
• This formula shows that centripetal acceleration is directly proportional to both the square of the angular velocity and the circle's radius.

As demonstrated in the exercise, differentiating the velocity vector with respect to time provides the acceleration vector \(\mathbf{a}(t)\), which confirms that the magnitude of the acceleration is indeed \( R \omega^2 \). The direction opposite to the position vector \(\mathbf{r}(t)\) further indicates its centripetal nature, oriented towards the center of the circle, ensuring the circular path is maintained.

Angular Velocity

Angular velocity \(\omega\) is a measure of the rotation rate and plays a fundamental role in describing circular motion. It represents how quickly an object moves through an angle and is usually expressed in radians per second. In the context of uniform circular motion, angular velocity is constant, simplifying many equations and calculations involved.
To better understand angular velocity's impact:

• It is the rate at which the particle moves around the circle; mathematically, \(\omega = \frac{\Delta \theta}{\Delta t}\), where \(\Delta \theta\) is the change in angle over a time interval \(\Delta t\).
• Angular velocity directly affects the tangential speed of the particle by the relation \(v = R \omega\), where \(v\) is the magnitude of the velocity.

In uniform circular motion, since \(\omega\) is constant, both the speed and the directional change are uniform, leading to the steady centripetal acceleration discussed earlier. This consistency in angular velocity makes it a key parameter when analyzing or predicting the motion's behavior, as seen in the current problem setup.