Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 27

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 -kg acrobat is located at \((3.00 \mathrm{~m}, 4.00 \mathrm{~m})\), and a 40 - \(\mathrm{kg}\) 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

Expert verified
Answer: The 20 kg acrobat must be positioned at coordinate (-3 m, 2 m).

Step by step solution

01

Find the center of mass for the system of three#acrobats

The formula for the center of mass in two-dimensional Cartesian coordinates is given by: $$ (\overline{x},\overline{y}) = (\frac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3}, \frac{m_1y_1+m_2y_2+m_3y_3}{m_1+m_2+m_3}) $$ Here, \((x_i, y_i)\) are the coordinates of the \(i\)-th acrobat and \(m_i\) is their mass.
02

Plug in the given values and set the system's center of mass to (0, 0)

We are given the masses and positions of the first two acrobats, and the mass of the third acrobat, so we can write the center of mass equation as: $$ (0, 0) = (\frac{30(3)+40(-2)+20x}{30+40+20}, \frac{30(4)+40(-2)+20y}{30+40+20}) $$
03

Solve the system of equations for x and y

We now have the following system of equations: $$ 0 = \frac{30(3)+40(-2)+20x}{30+40+20} $$ $$ 0 = \frac{30(4)+40(-2)+20y}{30+40+20} $$ To solve the system, let's multiply both sides of each equation by the denominator (90), then isolate x and y. First equation: $$ 0 = 30(3)+40(-2)+20x $$ $$ 20x = (-60) $$ $$ x = -3 $$ Second equation: $$ 0 = 30(4)+40(-2)+20y $$ $$ 20y= (120-80) $$ $$ y = 2 $$
04

Write the final answer

The location where the 20 kg acrobat must be positioned for the center of mass to be at the origin is at coordinate \((-3 \mathrm{~m}, 2 \mathrm{~m})\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A man with a mass of 55 kg stands up in a \(65-\mathrm{kg}\) canoe of length \(4.0 \mathrm{~m}\) floating on water. He walks from a point \(0.75 \mathrm{~m}\) from the back of the canoe to a point 0.75 m from the front of the canoe. Assume negligible friction between the canoe and the water. How far does the canoe move?

A 350 -kg cannon, sliding freely on a frictionless horizontal plane at a speed of \(7.5 \mathrm{~m} / \mathrm{s}\), shoots a 15 -kg cannonball at an angle of \(55^{\circ}\) above the horizontal. The velocity of the ball relative to the cannon is such that when the shot occurs, the cannon stops cold. What is the velocity of the ball relative to the cannon?

A student with a mass of \(40.0 \mathrm{~kg}\) can throw a \(5.00-\mathrm{kg}\) ball with a relative speed of \(10.0 \mathrm{~m} / \mathrm{s}\). The student is standing at rest on a cart of mass \(10.0 \mathrm{~kg}\) that can move without friction. If the student throws the ball horizontally, what will the velocity of the ball with respect to the ground be?

The Saturn \(V\) rocket, which was used to launch the Apollo spacecraft on their way to the Moon, has an initial mass \(M_{0}=2.80 \cdot 10^{6} \mathrm{~kg}\) and a final mass \(M_{1}=8.00 \cdot 10^{5} \mathrm{~kg}\) and burns fuel at a constant rate for \(160 .\) s. The speed of the exhaust relative to the rocket is about \(v=2700 . \mathrm{m} / \mathrm{s}\). a) Find the upward acceleration of the rocket, as it lifts off the launch pad (while its mass is the initial mass). b) Find the upward acceleration of the rocket, just as it finishes burning its fuel (when its mass is the final mass). c) If the same rocket were fired in deep space, where there is negligible gravitational force, what would be the net change in the speed of the rocket during the time it was burning fuel?

A circular pizza of radius \(R\) has a circular piece of radius \(R / 4\) removed from one side, as shown in the figure. Where is the center of mass of the pizza with the hole in it?
See all solutions

Recommended explanations on Physics Textbooks

Medical Physics

Read Explanation

Radiation

Read Explanation

Fundamentals of Physics

Read Explanation

Astrophysics

Read Explanation

Wave Optics

Read Explanation

Oscillations

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free