Question

Young acrobats are standing still on a circular horizontal platform suspended at the center. The origin of the two-dimensional Cartesian coordinate system is assumed to be at the center of the platform. A \(30.0-\mathrm{kg}\) acrobat is located at \((3.00 \mathrm{~m}, 4.00 \mathrm{~m})\), and a 40.0 -kg acrobat is located at \((-2.00 \mathrm{~m},-2.00 \mathrm{~m})\). Assuming that the acrobats stand still in their positions, where must a 20.0 -kg acrobat be located so that the center of mass of the system consisting of the three acrobats is at the origin and the platform is balanced?

Short answer

Answer: The third acrobat must be located at (-0.5 m, -2.0 m) for the center of mass to be at the origin, and for the platform to be balanced.

Step-by-step solution

Define the equation for the center of mass

The center of mass of a system of particles can be found by dividing the sum of the moments (mass times position) by the total mass of the system. For this problem, we have: \(x_{cm} = \frac{\sum m_ix_i}{\sum m_i}\) and \(y_{cm} = \frac{\sum m_iy_i}{\sum m_i}\) Where \(m_i\) and \((x_i, y_i)\) represent the mass and position of each acrobat, respectively.

Set the center of mass equal to the origin

Since we want the center of mass to be at the origin, we have: \(x_{cm} = 0\) and \(y_{cm} = 0\)

Define the given variables

We are given the positions and masses of two acrobats and the mass of the third acrobat: \(m_1 = 30.0\mathrm{~kg}\), \((x_1, y_1) = (3.00\mathrm{~m}, 4.00\mathrm{~m})\) \(m_2 = 40.0\mathrm{~kg}\), \((x_2, y_2) = (-2.00\mathrm{~m}, -2.00\mathrm{~m})\) \(m_3 = 20.0\mathrm{~kg}\), \((x_3, y_3) = (x, y)\) (unknown)

Set up equations for x and y

We will now substitute the given variables into the center of mass equations and set them equal to zero: \(0 = \frac{m_1x_1 + m_2x_2 + m_3x_3}{m_1 + m_2 + m_3}\) \(0 = \frac{m_1y_1 + m_2y_2 + m_3y_3}{m_1 + m_2 + m_3}\)

Solve for x and y coordinates of the 20.0-kg acrobat

Multiply both sides of each equation by the sum of the masses, \(m_1 + m_2 + m_3\), and solve for \(x_3\) and \(y_3\): \(0 = m_1x_1 + m_2x_2 + m_3x_3\) \(x_3 = \frac{-m_1x_1 - m_2x_2}{m_3}\) \(0 = m_1y_1 + m_2y_2 + m_3y_3\) \(y_3 = \frac{-m_1y_1 - m_2y_2}{m_3}\) Finally, substituting the given values: \(x_3 = \frac{-30(3) - 40(-2)}{20} = \frac{-90 + 80}{20} = \frac{-10}{20} = -0.5\mathrm{~m}\) \(y_3 = \frac{-30(4) - 40(-2)}{20} = \frac{-120 + 80}{20} = \frac{-40}{20} = -2.0\mathrm{~m}\) So, the 20.0-kg acrobat must be located at \((-0.5\mathrm{~m}, -2.0\mathrm{~m})\) in order for the center of mass of the system to be at the origin and the platform to be balanced.

Key concepts

Physics Problem Solving

When tackling problems in physics, it's crucial to approach them with a systematic method. Problem solving in physics often requires an understanding of various principles and the ability to apply them to different scenarios. The key steps in physics problem solving include identifying the relevant concepts, visualizing the problem, setting up equations, and carefully carrying out calculations.

For students trying to understand the concept of center of mass, visualization is particularly helpful. Imagine a balance point where an object - or system of objects, in this case, acrobats on a platform - can be perfectly balanced. By breaking down the problem into smaller parts, and identifying that we need to calculate the balancing point of the system, the solution becomes more approachable.

Think of each step as a building block towards the solution, and ensure you understand each phase before moving to the next. Understanding the meaning behind the mathematical expressions, like those used to calculate the center of mass, is essential for grasping the concept fully.

Equivant System

The concept of an equivalent system in physics refers to different systems that produce the same effect. In the context of center of mass calculations, we consider the system of acrobats as point masses located at their respective coordinates in the Cartesian system.

In our scenario, we aim for an 'equivalent system' for which the overall effect - the center of mass - is at the origin. This concept simplifies complex problems by reducing them to a more manageable form, without altering the fundamental physics involved. Here, individual acrobats represent point masses, and the question is essentially reduced to finding the correct position for the third point mass, so that the center of mass is at the desired location.

This kind of abstraction, replacing a complex system with a simpler but equivalent one, is a powerful tool in problem solving. For students, it's important to grasp that while the system becomes simpler in a mathematical sense, it retains all the essential properties of the original scenario needed to find the solution.

Cartesian Coordinates

Cartesian coordinates are a cornerstone of many physics problems as they provide a straightforward method to represent positions in space. In the case of the acrobats, their locations on the circular platform are denoted using Cartesian coordinates. This two-dimensional coordinate system uses an ordered pair of numbers \(x,y\), which represent the horizontal and vertical distances from the origin.

Students must understand how to plot points using these coordinates, and how to interpret the positions and movements of objects within this system. It is also crucial to understand the signs of these coordinates, as they indicate the direction from the origin: positive values indicate a position to the right/upward from the origin, while negative values represent a position to the left/downward.

With grasp of Cartesian coordinates, students can then solve problems such as locating the third acrobat in our example by understanding where \(x,y\) will have to place them in relation to the center of the platform to achieve balance. In physics, having this spatial understanding is indispensable in the analysis of forces, motion, and, as in our case, the center of mass.