Problem 23 - Advanced Engineering Mathematics

Question

Suppose that \(\mathbf{r}(t)=r_{0} \cos \omega t \mathbf{i}+r_{0} \sin \omega t \mathbf{j}\) is the position vector of an object that is moving in a circle of radius \(r_{0}\) in the \(x y\) -plane. If \(\|\mathbf{v}(t)\|=v\), show that the magnitude of the centripetal acceleration is \(a=\|\mathbf{a}(t)\|=v^{2} / r_{0}\).

Step-by-Step Solution

Answer

The centripetal acceleration magnitude is \( \frac{v^2}{r_0} \).

Step 1: Differentiate the Position Vector

First, we need to find the velocity vector \( \mathbf{v}(t) \) by differentiating the position vector \( \mathbf{r}(t) \) with respect to time \( t \). Given:\[ \mathbf{r}(t) = r_{0} \cos(\omega t) \mathbf{i} + r_{0} \sin(\omega t) \mathbf{j} \]Differentiate each component separately with respect to \( t \):- Derivative of \( r_{0}\cos(\omega t) \) is \( -r_{0}\omega \sin(\omega t) \)- Derivative of \( r_{0}\sin(\omega t) \) is \( r_{0}\omega \cos(\omega t) \)So, the velocity vector is:\[ \mathbf{v}(t) = -r_{0} \omega \sin(\omega t) \mathbf{i} + r_{0} \omega \cos(\omega t) \mathbf{j} \]

Step 2: Magnitude of Velocity

The magnitude of the velocity vector \( \|\mathbf{v}(t)\| \) is given to be a constant value \( v \). First confirm this assumption:Calculate \( \|\mathbf{v}(t)\| \):\[ \|\mathbf{v}(t)\| = \sqrt{(-r_{0}\omega \sin(\omega t))^2 + (r_{0}\omega \cos(\omega t))^2 } \]This simplifies to: \[ \|\mathbf{v}(t)\| = \sqrt{r_{0}^2 \omega^2 (\sin^2(\omega t) + \cos^2(\omega t)) } = r_{0} \omega \]Given \( \|\mathbf{v}(t)\| = v \), we equate and solve for \( \omega \):\[ v = r_{0} \omega \quad \Rightarrow \quad \omega = \frac{v}{r_{0}} \]

Step 3: Differentiate the Velocity Vector

Next, differentiate \( \mathbf{v}(t) \) with respect to time to find the acceleration vector \( \mathbf{a}(t) \).Given:\[ \mathbf{v}(t) = -r_{0} \omega \sin(\omega t) \mathbf{i} + r_{0} \omega \cos(\omega t) \mathbf{j} \]Differentiate each component:- Derivative of \( -r_{0}\omega \sin(\omega t) \) is \( -r_{0}\omega^2 \cos(\omega t) \)- Derivative of \( r_{0}\omega \cos(\omega t) \) is \( -r_{0}\omega^2 \sin(\omega t) \)So, the acceleration vector is:\[ \mathbf{a}(t) = -r_{0} \omega^2 \cos(\omega t) \mathbf{i} - r_{0} \omega^2 \sin(\omega t) \mathbf{j} \]

Step 4: Magnitude of Acceleration

Finally, calculate the magnitude of the acceleration vector \( \|\mathbf{a}(t)\| \).Calculate \( \|\mathbf{a}(t)\| \):\[ \|\mathbf{a}(t)\| = \sqrt{(-r_{0} \omega^2 \cos(\omega t))^2 + (-r_{0} \omega^2 \sin(\omega t))^2 } \]This simplifies using trigonometric identity \( \cos^2(\omega t) + \sin^2(\omega t) = 1 \):\[ \|\mathbf{a}(t)\| = \sqrt{r_{0}^2 \omega^4} = r_{0} \omega^2 \]Substitute \( \omega = \frac{v}{r_{0}} \) from Step 2:\[ r_{0} \omega^2 = r_{0} \left(\frac{v}{r_{0}}\right)^2 = \frac{v^2}{r_{0}} \]

Step 5: Conclusion

Since the magnitude of the acceleration is \( \|\mathbf{a}(t)\| = \frac{v^2}{r_{0}} \), this confirms that the centripetal acceleration indeed has the magnitude \( a = \frac{v^2}{r_{0}} \), as required.

Key Concepts

Position Vector

In circular motion, an object's path and location at any moment can be represented by a position vector. For an object moving in a circle of radius \( r_0 \) in the xy-plane, the position vector \( \mathbf{r}(t) \) is expressed by the equations \( r_{0} \cos(\omega t) \mathbf{i} + r_{0} \sin(\omega t) \mathbf{j} \). These equations allow us to capture the object's location as a function of time.
The components \( r_{0}\cos(\omega t) \) and \( r_{0}\sin(\omega t) \) correspond to the x and y coordinates, respectively, based on trigonometric functions. The term \( \omega \) represents the angular velocity, describing how fast the object is moving around the circle.
To visualize this, imagine the object moving counterclockwise around the circle. As time progresses, the cosine component influences the horizontal position, while the sine component affects the vertical position. Together, these components ensure the position vector maintains the circular path defined by the constant radius \( r_0 \).

Differentiation

Differentiation is crucial in motion analysis, helping to derive both velocity and acceleration from a position vector. By differentiating the position vector \( \mathbf{r}(t) \) with respect to time \( t \), we obtain the velocity vector \( \mathbf{v}(t) \). Differentiating again, produces the acceleration vector \( \mathbf{a}(t) \).
Starting with position \( \mathbf{r}(t) = r_{0}\cos(\omega t) \mathbf{i} + r_{0}\sin(\omega t) \mathbf{j} \), we differentiate:

• Derivative of \( r_0 \cos(\omega t) \) yields \( -r_0 \omega \sin(\omega t) \).
• Derivative of \( r_0 \sin(\omega t) \) gives \( r_0 \omega \cos(\omega t) \).

These derivatives form the velocity vector \( \mathbf{v}(t) = -r_0 \omega \sin(\omega t) \mathbf{i} + r_0 \omega \cos(\omega t) \mathbf{j} \).
Differentiating the velocity vector leads to the acceleration vector, shown as:

• Derivative of \( -r_0 \omega \sin(\omega t) \) becomes \( -r_0 \omega^2 \cos(\omega t) \).
• Derivative of \( r_0 \omega \cos(\omega t) \) results in \( -r_0 \omega^2 \sin(\omega t) \).

Thus, \( \mathbf{a}(t) = -r_0 \omega^2 \cos(\omega t) \mathbf{i} - r_0 \omega^2 \sin(\omega t) \mathbf{j} \), illustrating how differentiation reveals deeper motion insights.

Magnitude of Acceleration

The magnitude of an acceleration vector is essential to understand how the speed of an object in motion changes. In circular motion, this concept is often linked with centripetal acceleration, which keeps the object traveling in a circular path.
Given the acceleration vector \( \mathbf{a}(t) = -r_0 \omega^2 \cos(\omega t) \mathbf{i} - r_0 \omega^2 \sin(\omega t) \mathbf{j} \), we find its magnitude by calculating:\[\|\mathbf{a}(t)\| = \sqrt{(-r_0 \omega^2 \cos(\omega t))^2 + (-r_0 \omega^2 \sin(\omega t))^2}\]Utilizing the well-known trigonometric identity \( \cos^2(\omega t) + \sin^2(\omega t) = 1 \), it simplifies to:\[\|\mathbf{a}(t)\| = \sqrt{r_0^2 \omega^4} = r_0 \omega^2\]Substituting \( \omega = \frac{v}{r_0} \), as derived from equating velocity magnitudes, into the equation:\[\|\mathbf{a}(t)\| = r_0 \left(\frac{v}{r_0}\right)^2 = \frac{v^2}{r_0}\]This result confirms the magnitude of centripetal acceleration \( a \), ensuring it aligns with the expression \( a = \frac{v^2}{r_0} \), crucial in the context of circular motion. This insight into acceleration magnitude helps solidify the relationship between velocity, radius, and centripetal forces ensuring circular motion.