Question

Consider the motion of a point (or particle) on the circumference of a rolling circle. As the circle rolls, it generates the cycloid \(\mathbf{r}(t)=b(\omega t-\sin \omega t) \mathbf{i}+b(1-\cos \omega t) \mathbf{j}\) where \(\omega\) is the constant angular velocity of the circle and \(b\) is the radius of the circle. Find the velocity and acceleration vectors of the particle. Use the results to determine the times at which the speed of the particle will be (a) zero and (b) maximized.

Short answer

The velocity vector of the particle is \(v(t) = b\omega(1 - \cos(\omega t))\mathbf{i} + b\omega \sin(\omega t)\mathbf{j}\) and the acceleration vector is \(a(t) = \omega^2 b \sin(\omega t)\mathbf{i} + \omega^2 b \cos(\omega t)\mathbf{j}\). The speed of the particle will be maximum at times \(t = \frac{\pi}{2\omega}+ \frac{n\pi}{\omega}\) and zero at times \(t = \frac{n\pi}{\omega}\) where n is any integer.

Step-by-step solution

Find the velocity vector

The velocity vector \(v(t)\) of the particle can be found by differentiating the position vector \(r(t)\). To do this, differentiate the position function for each component i and j separately.\n\nTo find \(v(t)\), let's differentiate \(r(t)\) to get:\(v(t) = \frac{dr}{dt} = b\omega(1 - \cos(\omega t))\mathbf{i} + b\omega \sin(\omega t)\mathbf{j}\)

Find the acceleration vector

The acceleration vector \(a(t)\) of the particle can be found by differentiating the velocity vector \(v(t)\). Again, differentiate the velocity function for each component i and j separately.\n\nTo find \(a(t)\), let's differentiate \(v(t)\) to get:\(a(t) = \frac{dv}{dt} = \omega^2 b \sin(\omega t)\mathbf{i} + \omega^2 b \cos(\omega t)\mathbf{j}\)

Find when the speed is zero or at its maximum

The magnitude of the velocity vector, or the speed \(s(t)\), is \(s(t) = |v(t)| = \sqrt{(v_x(t))^2+(v_y(t))^2}\). This comes out to be \(s(t) = b\omega |\sin(\omega t)|\) . Now, the speed of the particle will be maximum when \(|\sin(\omega t)| = 1\), which implies \(\omega t = \pi/2 + n\pi\) where n is any integer. The speed will be zero when \(|\sin(\omega t)| = 0\), which implies \(\omega t = n\pi\), again where n is any integer.

Key concepts

Velocity Vector Calculus

In the study of a particle's movement, such as the cycloid motion, it is crucial to understand its velocity at any given moment. The velocity vector represents the rate of change of the particle's position over time and indicates both speed and direction. To find this, calculus is employed by differentiating the position vector function with respect to time.

For a particle on a cycloid path described by the position vector \(\mathbf{r}(t)=b(\omega t-\sin \omega t) \mathbf{i}+b(1-\cos \omega t) \mathbf{j}\), where \(\omega\) is the angular velocity and \(b\) the radius of the rolling circle, we differentiate each component to find the velocity vector \(v(t)\).

By applying calculus, the x-component corresponds to the horizontal motion and is given by \(b\omega(1 - \cos(\omega t))\), and the y-component, representing vertical motion, is \(b\omega \sin(\omega t)\). Thus, the velocity vector \(v(t)\) is calculated as:
\[v(t) = \frac{dr}{dt} = b\omega(1 - \cos(\omega t))\mathbf{i} + b\omega \sin(\omega t)\mathbf{j}\].

Understanding this process is essential for analyzing the characteristics of the particle's motion, such as its speed at different instances.

Acceleration Vector Calculus

After determining the velocity of a particle, uncovering its acceleration is the next step to fully understand its motion. The acceleration vector is found by differentiating the velocity vector with respect to time. This vector highlights how the velocity of the particle changes over time, reflecting the forces acting on it.

In our cycloid example, where the velocity vector is given by \(v(t) = b\omega(1 - \cos(\omega t))\mathbf{i} + b\omega \sin(\omega t)\mathbf{j}\), we differentiate each component with respect to time to get the acceleration. Differentiating the x-component and the y-component of \(v(t)\) yields:
\[a(t) = \frac{dv}{dt} = \omega^2 b \sin(\omega t)\mathbf{i} + \omega^2 b \cos(\omega t)\mathbf{j}\].

This calculus process allows us to see that the acceleration vector can have alternating signs depending on \(\omega t\), showing the cyclical nature of the forces exerted on the particle as it follows the cycloid path. Having both velocity and acceleration vectors enables us to predict the particle's future position and behavior.

Particle Speed Maximization

An exciting aspect of studying motion is finding when the speed of a particle is maximized. Speed, as a scalar quantity, refers to how fast a particle is moving irrespective of its direction. To maximize this quantity, it is necessary to first determine an expression for speed by finding the magnitude of the velocity vector.

For the cycloid motion in our problem, we derive the speed \(s(t)\) as \(s(t) = |v(t)| = \sqrt{(v_x(t))^2+(v_y(t))^2}\). By plugging in the components from the velocity vector we found earlier, we attain the expression for speed as a function of time: \(s(t) = b\omega |\sin(\omega t)|\).

Maximization occurs when \(s(t)\) is at its greatest value, which means that \(|\sin(\omega t)|\) should be maximized. Since the maximum value of the sine function is 1, the particle speed is maximum when \(\omega t = \frac{\pi}{2} + n\pi\), where \(n\) is any integer. Notably, understanding this concept helps in scenarios where maximum speed is desired, such as planning the optimal path for a roller coaster. Insights into speed variations along a path can lead to more efficient designs in engineering and other practical applications.