Problem 68 - Fundamentals of Physics

Question

A cat rides a merry-go-round turning with uniform circular motion. At time \(t_{1}=2.00 \mathrm{~s}\), the cat's velocity is \(\vec{v}_{1}=\) \((3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(4.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\), measured on a horizontal \(x y\) coordinate system. At \(t_{2}=5.00 \mathrm{~s}\), the cat's velocity is \(\vec{v}_{2}=(-3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+\) \((-4.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\). What are (a) the magnitude of the cat's centripetal acceleration and (b) the cat's average acceleration during the time interval \(t_{2}-t_{1}\), which is less than one period?

Step-by-Step Solution

Answer

The cat's centripetal acceleration is approximately 3.33 m/s², and its average acceleration is -2.00 i - 2.67 j m/s².

Step 1: Calculate the magnitude of the velocity

To find the magnitude of the initial velocity \(\vec{v}_1\) and the final velocity \(\vec{v}_2\), we use the formula for magnitude: \[ |\vec{v}| = \sqrt{v_x^2 + v_y^2} \]Substitute for \(\vec{v}_1\):\[ |\vec{v}_1| = \sqrt{(3.00)^2 + (4.00)^2} = 5.00 \, \text{m/s} \]Similarly, since \(\vec{v}_2\) is \((-3.00 \, \text{m/s}) \hat{i} + (-4.00 \, \text{m/s}) \hat{j}\):\[ |\vec{v}_2| = \sqrt{(-3.00)^2 + (-4.00)^2} = 5.00 \, \text{m/s} \]

Step 2: Determine the change in velocity

To find the change in velocity \(\Delta \vec{v}\), subtract the initial velocity from the final velocity:\[ \Delta \vec{v} = \vec{v}_2 - \vec{v}_1 \]Substituting the given values:\[ \Delta \vec{v} = (-3.00 \, \hat{i} - 4.00 \, \hat{j}) - (3.00 \, \hat{i} + 4.00 \, \hat{j}) \ = [-3.00 - 3.00] \, \hat{i} + [-4.00 - 4.00] \, \hat{j} \ = -6.00 \, \hat{i} - 8.00 \, \hat{j} \, \text{m/s} \]

Step 3: Calculate the magnitude of the change in velocity

Using the magnitude formula again:\[ |\Delta \vec{v}| = \sqrt{(-6.00)^2 + (-8.00)^2} \]\[ |\Delta \vec{v}| = \sqrt{36 + 64} = \sqrt{100} = 10.00 \, \text{m/s} \]

Step 4: Calculate the centripetal acceleration

Centripetal acceleration \(a_c\) is constant for uniform circular motion and given by:\[ a_c = \frac{v^2}{r} \]The radius \(r\) does not need to be calculated directly because it cancels when using total change in angular displacement over a period time, thus:\[ a_c = \frac{|\Delta \vec{v}|}{\Delta t} \]Where \(\Delta t = t_2 - t_1 = 5.00 - 2.00 = 3.00\, \text{s} \).\[ a_c = \frac{10.00}{3.00} \approx 3.33 \, \text{m/s}^2 \]

Step 5: Calculate average acceleration

Average acceleration \(\vec{a}_{avg}\) is the change in velocity divided by the time interval:\[ \vec{a}_{avg} = \frac{\Delta \vec{v}}{\Delta t} \]Substituting the values:\[ \vec{a}_{avg} = \frac{-6.00 \, \hat{i} - 8.00 \, \hat{j}}{3.00} \]Divide each component by 3:\[ \vec{a}_{avg} = (-2.00 \, \hat{i} - 2.67 \, \hat{j}) \, \text{m/s}^2 \]

Key Concepts

Centripetal Acceleration

Centripetal acceleration is a crucial concept when studying circular motion. It describes the net acceleration that acts on an object moving in a circular path, always directed towards the center of the circle. This inward acceleration is what keeps the object moving in a circle, rather than flying off in a straight line. In our given exercise, the cat on the merry-go-round experiences centripetal acceleration as it moves uniformly in a circular path.
To calculate centripetal acceleration, we use the formula \[ a_c = \frac{v^2}{r} \]where \(v\) is the velocity of the object and \(r\) is the radius of the circular path. Often, in calculations for problems like these, the initial steps may focus on finding the change in velocity and the time interval over which it occurs. However, direct usage of radius isn't required if sufficient information is given about velocity changes and their intervals.
For the cat on the merry-go-round, this acceleration is essential to maintaining its circular motion, ensuring it continues moving round and round rather than veering off in another direction. Understanding this concept helps explain not only amusement park rides but also the orbits of planets and satellites.

Velocity Vector

Velocity vectors provide valuable information about both the speed and direction of an object's movement. In the context of uniform circular motion, these vectors are constantly changing direction even though the object's speed might remain constant. This particular problem involving the cat demonstrates how the velocity vector changes across different points in time.
The initial velocity \(\vec{v}_1 = (3.00 \, \text{m/s}) \hat{i} + (4.00 \, \text{m/s}) \hat{j}\) and final velocity \(\vec{v}_2 = (-3.00 \, \text{m/s}) \hat{i} + (-4.00 \, \text{m/s}) \hat{j}\) show how the direction of the velocity vector flips completely. Using the formula for the magnitude of a vector, \[ |\vec{v}| = \sqrt{v_x^2 + v_y^2} \]one can see that while magnitude might stay the same (as in this exercise), the direction, represented by the unit vectors \(\hat{i}\) and \(\hat{j}\), can change.
This variation in direction is what gives rise to centripetal acceleration and affects the motion around the circle. By understanding velocity vectors thoroughly, one can better predict and analyze the movement of objects in circular paths.

Average Acceleration

The concept of average acceleration is key when analyzing changes in velocity over a specific time period. Defined as the change in velocity divided by the time over which the change occurs, average acceleration gives a simple snapshot of how quickly an object is speeding up or slowing down. In our merry-go-round scenario, we're interested in the cat's acceleration between two specific points in time, \(t_1\) and \(t_2\).
The average acceleration formula is straightforward:\[ \vec{a}_{\text{avg}} = \frac{\Delta \vec{v}}{\Delta t} \]Inserting the given values, where \(\Delta \vec{v} = -6.00 \, \hat{i} - 8.00 \, \hat{j} \text{ m/s}\) and \(\Delta t = 3.00\) seconds, allows us to compute:\[ \vec{a}_{\text{avg}} = (-2.00 \, \hat{i} - 2.67 \, \hat{j}) \, \text{m/s}^2 \]Breaking this calculation into components makes it easier to digest. This allows students to see how each directional change contributes to overall movement and acceleration.
Average acceleration is a useful measure for understanding motion over intervals, especially in situations where the motion involves change over a circular path or when velocities change significantly. For students, mastering this concept rounds out a solid understanding of basic acceleration scenarios.