Problem 57 - Fundamentals of Physics

Question

A carnival merry-go-round rotates about a vertical axis at a constant rate. A man standing on the edge has a constant speed of \(3.66 \mathrm{~m} / \mathrm{s}\) and a centripetal acceleration \(\vec{a}\) of magnitude \(1.83 \mathrm{~m} / \mathrm{s}^{2}\) Position vector \(\vec{r}\) locates him relative to the rotation axis. (a) What is the magnitude of \(\vec{r}\) ? What is the direction of \(\vec{r}\) when \(\vec{a}\) is directed (b) due east and (c) due south?

Step-by-Step Solution

Answer

(a) \(r \approx 7.32 \, \text{m}\). (b) \(\vec{r}\) points west. (c) \(\vec{r}\) points north.

Step 1: Understand the Relationship Between Centripetal Acceleration and Radius

Centripetal acceleration \(a_c\) is given by the formula \(a_c = \frac{v^2}{r}\), where \(v\) is the tangential speed and \(r\) is the radius of the circular path. In this problem, \(v = 3.66\, \text{m/s}\) and \(a_c = 1.83\, \text{m/s}^2\). We can rearrange the formula to solve for \(r\): \(r = \frac{v^2}{a_c}\).

Step 2: Calculate the Magnitude of \(\vec{r}\)

Using the formula \(r = \frac{v^2}{a_c}\), substitute \(v = 3.66\, \text{m/s}\) and \(a_c = 1.83\, \text{m/s}^2\):\[r = \frac{(3.66)^2}{1.83} = \frac{13.3956}{1.83} \approx 7.32 \, \text{m}.\]

Step 3: Determine the Direction of \(\vec{r}\) When \(\vec{a}\) is Due East

Centripetal acceleration is always directed towards the center of the rotation. Hence, if \(\vec{a}\) is directed due east, the direction of \(\vec{r}\) is directly opposite to \(\vec{a}\), which would be towards the west.

Step 4: Determine the Direction of \(\vec{r}\) When \(\vec{a}\) is Due South

Similar to Step 3, if \(\vec{a}\) is directed due south, the direction of \(\vec{r}\) will be opposite to \(\vec{a}\), meaning \(\vec{r}\) points towards the north.

Key Concepts

Centripetal Acceleration

Centripetal acceleration is a crucial concept in understanding how objects move along a circular path. It is defined as the acceleration that directs an object towards the center of a circular path it is traversing. Unlike typical linear acceleration, centripetal acceleration does not change the speed of the object but alters its direction of motion. This type of acceleration arises because the direction of velocity is constantly changing as the object moves in a circle, even if its speed remains constant.

The formula for calculating centripetal acceleration is given by \(a_c = \frac{v^2}{r}\), where \(v\) represents the constant speed of the object, and \(r\) is the radius of the circular path. Here, the square of the speed is divided by the radius, indicating the dependence of centripetal acceleration on both these parameters. Keep in mind that the larger the radius, the smaller the acceleration required for the same speed. This relationship is vital when figuring out scenarios like the one in the merry-go-round problem, where knowing either the speed or acceleration can help determine the other parameters.

Circular Motion

In physics, circular motion refers to the motion of a body along a circular path. It is a central theme in mechanics since it illustrates how forces can cause an object to move in a curved trajectory. Within circular motion, there are several key ideas and formulas that help in understanding the dynamics involved.

When an object moves with a constant speed in a circular path, it experiences uniform circular motion. Let's assume we have a situation like a rider on a merry-go-round, where their speed is constant, but their direction keeps changing. This change in direction is where centripetal acceleration comes into play, constantly pulling the rider towards the center, maintaining the circular path.

The importance of centripetal force, which results from the centripetal acceleration, cannot be overstated. In essence, centripetal force is what keeps an object in circular motion rather than allowing it to move off in a tangent. It can be further explored by the equation \(F_c = m \cdot a_c\), where \(m\) is the mass of the rotating object and \(a_c\) is the centripetal acceleration.

Physics Problem Solving

When facing physics problems involving circular motion, successful problem-solving involves understanding the relationships and applying correct formulas. Effective problem-solving often starts with clearly defining the known quantities and what needs to be determined. In the case of centripetal motion problems, identifying the correct formula for centripetal acceleration or force is the first step.

Many circular motion problems require an understanding of formulas like \(a_c = \frac{v^2}{r}\). Rearranging this formula based on the information given can help in finding missing variables, such as radius or speed. It's important to remember to keep units consistent throughout the calculation to avoid errors.

When considering the problem of determining the direction of vectors like position or acceleration, it's crucial to comprehend the geometry of the circle involved. For instance, if you know the direction of the acceleration vector, the position vector will conventionally point directly opposite to it on the circular path. This understanding aids in visualizing and solving the problem correctly.