Problem 165 - Calculus Volume 3

Question

Consider the motion of a point on the circumference of a rolling circle. As the circle rolls, it generates the cycloid \(\mathbf{r}(t)=(\omega t-\sin (\omega t) \mid \mathbf{i}+(1-\cos (\omega t)) \mathbf{j}\), where \(\omega\) is the angular velocity of the circle and \(b\) is the radius of the circle: Find the equations for the velocity, acceleration, and speed of the particle at any time.

Step-by-Step Solution

Answer

Velocity: \( \mathbf{v}(t) = (\omega - \omega \cos(\omega t)) \mathbf{i} + \omega \sin(\omega t) \mathbf{j} \). Acceleration: \( \mathbf{a}(t) = \omega^2 \sin(\omega t) \mathbf{i} + \omega^2 \cos(\omega t) \mathbf{j} \). Speed: \( 2\omega |\sin(\omega t/2)| \).

Step 1: Differentiate to Find Velocity

To find the velocity of the point, we differentiate the position vector \( \mathbf{r}(t) = (\omega t - \sin (\omega t))\mathbf{i} + (1 - \cos (\omega t))\mathbf{j} \) with respect to \( t \). The derivative of the first component is \( \omega - \omega \cos(\omega t) \), and the derivative of the second component is \( \omega \sin(\omega t) \). Thus, the velocity \( \mathbf{v}(t) = \left( \omega - \omega \cos(\omega t) \right) \mathbf{i} + \omega \sin(\omega t) \mathbf{j} \).

Step 2: Differentiate to Find Acceleration

To find the acceleration vector, differentiate the velocity vector \( \mathbf{v}(t) = (\omega - \omega \cos(\omega t)) \mathbf{i} + \omega \sin(\omega t) \mathbf{j} \) with respect to \( t \). The derivative of the first component is \( \omega^2 \sin(\omega t) \), and the derivative of the second component is \( \omega^2 \cos(\omega t) \). Therefore, the acceleration \( \mathbf{a}(t) = \omega^2 \sin(\omega t) \mathbf{i} + \omega^2 \cos(\omega t) \mathbf{j} \).

Step 3: Compute Speed from Velocity

The speed of the particle is the magnitude of the velocity vector \( \mathbf{v}(t) = (\omega - \omega \cos(\omega t)) \mathbf{i} + \omega \sin(\omega t) \mathbf{j} \). Calculate \( |\mathbf{v}(t)| = \sqrt{(\omega - \omega \cos(\omega t))^2 + (\omega \sin(\omega t))^2} \). Simplifying, we find that \( |\mathbf{v}(t)| = \sqrt{(\omega (1 - \cos(\omega t)))^2 + (\omega \sin(\omega t))^2} = \sqrt{2\omega^2 (1 - \cos(\omega t))} \). This simplifies to \( 2\omega |\sin(\omega t/2)| \), a known identity.

Key Concepts

Velocity Vector Calculus

When studying cycloid motion, one of the first steps is understanding velocity. The velocity vector describes how fast and in which direction a point is moving at any given time. To find the velocity vector for a point on the cycloid curve, we differentiate the position vector, \( \mathbf{r}(t) = (\omega t - \sin (\omega t))\mathbf{i} + (1 - \cos (\omega t))\mathbf{j} \), with respect to time \( t \).
The differentiation involves applying the chain rule to each component:

• The component \( \omega t - \sin(\omega t) \) differentiates to \( \omega - \omega \cos(\omega t) \).
• The component \( 1 - \cos(\omega t) \) differentiates to \( \omega \sin(\omega t) \).

Thus, the velocity vector is given by:
\[ \mathbf{v}(t) = (\omega - \omega \cos(\omega t)) \mathbf{i} + \omega \sin(\omega t) \mathbf{j} \]
This vector represents both the speed and direction of the particle at any instant along the cycloid.

Acceleration Vector Calculus

Once we have the velocity vector, the next step is to determine the acceleration vector, which tells us how the velocity itself changes over time. To find this, we differentiate the velocity vector \( \mathbf{v}(t) \) with respect to time \( t \).
By differentiating each component of the velocity vector, we derive the acceleration components:

• The term \( \omega - \omega \cos(\omega t) \) becomes \( \omega^2 \sin(\omega t) \).
• The term \( \omega \sin(\omega t) \) becomes \( \omega^2 \cos(\omega t) \).

Now, the acceleration vector can be expressed as:
\[ \mathbf{a}(t) = \omega^2 \sin(\omega t) \mathbf{i} + \omega^2 \cos(\omega t) \mathbf{j} \]
This vector indicates how the velocity vector changes its magnitude and direction, showing the dynamic behavior of a cycloidal motion.

Speed Calculation

The speed of an object in motion is the magnitude of its velocity vector. Calculating speed involves finding the length of the velocity vector.
For our cycloid, the velocity vector is \( \mathbf{v}(t) = (\omega - \omega \cos(\omega t)) \mathbf{i} + \omega \sin(\omega t) \mathbf{j} \).
The speed \( |\mathbf{v}(t)| \) is given by:

• First, square the individual components of the velocity and then add them:
  \( (\omega - \omega \cos(\omega t))^2 + (\omega \sin(\omega t))^2 \).
• This simplifies using trigonometric identities, resulting in \( 2\omega^2 (1 - \cos(\omega t)) \).

Thus, the speed can be further simplified to:
\[ |\mathbf{v}(t)| = 2\omega |\sin(\omega t/2)| \]
This expression effectively showcases how the speed varies with time as the point traverses the cycloid path.

Differentiation in Calculus

Differentiation is a key calculus tool used to analyze motion, such as cycloid motion. It allows us to calculate rates of change like velocity and acceleration from a given position vector.
To differentiate a function, apply derivative rules appropriately:

• The derivative of a sum is the sum of the derivatives.
• The derivative of \( \sin(\omega t) \) is \( \omega \cos(\omega t) \).
• The derivative of \( \cos(\omega t) \) is \(-\omega \sin(\omega t) \).

Differentiation not only helps calculate velocity and acceleration but also assists in understanding how different motions are related. It highlights how small changes in position lead to corresponding changes in velocity and acceleration, providing insights into the dynamic aspects of movement in systems like a cycloid.