Question

An object moves in the \(x y\) plane such that \(x=R \cos \omega t\) and \(y=R \sin \omega t\). Here \(x\) and \(y\) are the coordinates of the object, is the time, and \(R\) and \(\omega\) are constants. (a) Eliminate \(t\) between these equations to find the equation of the curve in which the object moves. What is this curve? What is the meaning of the constant \(\omega ?(b)\) Differentiate the equations for \(x\) and \(y\) with respect to the time to find the \(x\) and \(y\) components of the velocity of the body, \(v_{x}\) and \(v_{y}\). Combine \(v_{x}\) and \(v_{y}\) to find the magnitude and direction of \(v .\) Describe the motion of the object. ( \(c\) ) Differentiate \(v_{x}\) and \(v_{y}\) with respect to the time to obtain the magnitude and direction of the resultant acceleration.

Short answer

The object travels in a circle of radius \(R\) with a uniform angular speed \(\omega\). The direction of velocity at any moment is tangential to the circle, while the direction of acceleration is always towards the centre of the circular path.

Step-by-step solution

Eliminate parameter and find the path

To eliminate \(t\), use the trigonometric identity \(\sin^2(t) + \cos^2(t) = 1\), which can be rearranged as \(\sin^2(t) = 1 - \cos^2(t)\). Substituting the equations \(x=R \cos \omega t\) and \(y=R \sin \omega t\), and eliminating \(t\), the equation becomes \(x^2 + y^2 = R^2\). Thus, the curve along which the object moves is a circle with radius \(R\).

Differentiate to find the components of velocity

The derivatives \(dx/dt\) and \(dy/dt\) will represent the x and y components of velocity (i.e., \(v_x\) and \(v_y\)) respectively. By differentiating, we obtain \(v_x = -\omega R \sin \omega t\) and \(v_y = \omega R \cos \omega t\).

Combine to find the magnitude and direction of velocity

The magnitude of velocity \(v\) is found by taking the square root of the sum of squares of the components, hence \(v = \sqrt{(v_x^2 + v_y^2)} = \sqrt{(\omega R)^2} = \omega R\). The direction is the angle \(\theta\) such that \(\tan \theta = v_y / v_x = - \cot \omega t\).

Differentiate velocity to find the components of acceleration

The derivatives \(dv_x/dt\) and \(dv_y/dt\) will represent the x and y components of acceleration. This results in \(a_x = -\omega^2 R \cos \omega t\) and \(a_y = -\omega^2 R \sin \omega t\).

Combine to find the magnitude and direction of acceleration

The magnitude of acceleration \(a\) is found by taking the square root of the sum of squares of the components, which yields \(a = \sqrt{(a_x^2 + a_y^2)} = \sqrt{(\omega^2 R)^2} = \omega^2 R\). The direction is the angle \(\phi\) such that \(\tan \phi = a_y / a_x = - \tan \omega t\).

Key concepts

Parametric Equations

Parametric equations are an elegant way to represent curves in the plane by using an independent parameter, typically time (\t)). In our exercise, the object's position is described by two equations: for the x-coordinate, \(x = R\cos\omega t\) and for the y-coordinate, \(y = R\sin\omega t\). These equations define the motion of the object in the plane as a function of time.

When plotting the path of this object, we aren't restricting ourselves to the typical function relationship of y depending on x. Instead, both x and y depend on \(t\). This independence means we can represent more complex paths, such as loops and circles – which in this case, after eliminating \(t\), the path is revealed to be a circle with radius \(R\). This form of motion expression allows for easier analysis of movement in physics, especially in circular motion.

Velocity and Acceleration

When it comes to circular motion, understanding velocity and acceleration is crucial. Velocity is the rate of change of an object's position with respect to time, while acceleration is the rate of change of velocity with respect to time. In the provided example, by differentiating the position parametric equations with respect to time, we obtain the components of velocity, \(v_x\) and \(v_y\).

The magnitude and direction of velocity are then determined by combining these components, illustrating the object's instantaneous speed and movement direction within its circular path. Similarly, further differentiation provides us the acceleration components, which when combined, give the total acceleration of the object. This reveals that the object undergoes a centripetal acceleration, always directed towards the center of the circular path.

Trigonometric Identities

Trigonometric identities like \(\sin^2(t) + \cos^2(t) = 1\) are key tools in simplifying the expressions in physics, particularly in circular motion problems. By applying this identity, we effectively eliminated the parameter \(t\) in our exercise, showing that the object's path is a circle, given by the equation \(x^2 + y^2 = R^2\).

These identities are not just mathematical curiosities; they come from the unit circle's fundamental properties and are critical in solving a variety of physics problems. They allow us to transition between different trigonometric functions seamlessly and to solve for unknowns in complex equations involving angles and lengths.

Differential Calculus in Physics

Differential calculus is a potent mathematical tool in physics, helping us analyze motion by finding rates of change. In the context of circular motion, calculus allows us to find velocity and acceleration as derivatives of position with respect to time. As we saw in the exercise, differentiating the parametric equations of position yields the velocity components. A further derivative gives us the acceleration components.

This calculus application did more than quantify change; it revealed the nature of the object's motion - in this case, it's uniform circular motion, as indicated by the constant speed and centripetal acceleration. In physics, these tools enable us to describe and predict objects' motion in different frameworks.