/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 76 A uniform rod of length \(30.0 \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 76

A uniform rod of length \(30.0 \mathrm{cm}\) is bent into the shape of an inverted U. Each of the three sides is of length \(10.0 \mathrm{cm} .\) Find the location, in \(x\) - and \(y\) -coordinates, of the CM as measured from the origin.

Short Answer

Expert verified
Answer: The center of mass (CM) for the bent rod can be found at coordinates (6.67 cm, 5 cm).

Step by step solution

01

Label the rods and find their individual centers of mass

First, let's label the three rods: Rod A is the left vertical side, Rod B is the horizontal top part, and Rod C is the right vertical side. As each rod is uniform, the center of mass of each rod is at its midpoint. Here are the CM coordinates for each rod: - For Rod A: CM_A = (0, 5 cm) - For Rod B: CM_B = (5 cm, 10 cm) - For Rod C: CM_C = (10 cm, 5 cm)
02

Calculate the moment of each individual rod's CM about the x- and y-axes

To find the overall center of mass of the system, we need to compute the moment of each rod's CM about the x and y-axes. The moment of a rod's center of mass about an axis is given by the product of the distance of its CM from the given axis and the mass of the rod. Let m be the mass per unit length of the uniform rod. Thus, the mass of each 10 cm rod is M = 10m. We can now compute the moments of each rod's CM as follows: - For Rod A: - Moment about x-axis: M * CM_A_y = 10m * 5 cm = 50m cm - Moment about y-axis: M * CM_A_x = 10m * 0 cm = 0m cm - For Rod B: - Moment about x-axis: M * CM_B_y = 10m * 10 cm = 100m cm - Moment about y-axis: M * CM_B_x = 10m * 5 cm = 50m cm - For Rod C: - Moment about x-axis: M * CM_C_y = 10m * 5 cm = 50m cm - Moment about y-axis: M * CM_C_x = 10m * 10 cm = 100m cm
03

Compute the total moment about the x- and y-axes

We can now add up the moments of each rod's CM for the total moment about x and y axes: - Total moment about x-axis: Moment_A_x + Moment_B_x + Moment_C_x = 50m + 100m + 50m = 200m cm - Total moment about y-axis: Moment_A_y + Moment_B_y + Moment_C_y = 0 + 50m + 100m = 150m cm
04

Calculate the overall center of mass of the system

The overall center of mass of the system is determined by using the moment balance equation: CM_x = (Total moment about x-axis) / (Total mass) CM_y = (Total moment about y-axis) / (Total mass) The total mass of the uniform rod is 30m (10m for each segment). Therefore, we can find the overall CM as follows: CM_x = (200m cm) / (30m) = 20/3 cm 鈮� 6.67 cm CM_y = (150m cm) / (30m) = 5 cm Finally, the location of the center of mass (CM) of the bent rod, expressed in x- and y-coordinates, is given by: CM = (6.67 cm, 5 cm)

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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

Two cars, each of mass \(1300 \mathrm{kg},\) are approaching each other on a head-on collision course. Each speedometer reads $19 \mathrm{m} / \mathrm{s} .$ What is the magnitude of the total momentum of the system?
A stationary 0.1-g fly encounters the windshield of a \(1000-\mathrm{kg}\) automobile traveling at \(100 \mathrm{km} / \mathrm{h} .\) (a) What is the change in momentum of the car due to the fly? (b) What is the change of momentum of the fly due to the car? (c) Approximately how many flies does it take to reduce the car's speed by \(1 \mathrm{km} / \mathrm{h} ?\)
In a circus trapeze act, two acrobats actually fly through the air and grab on to one another, then together grab a swinging bar. One acrobat, with a mass of \(60 \mathrm{kg},\) is moving at \(3.0 \mathrm{m} / \mathrm{s}\) at an angle of \(10^{\circ}\) above the horizontal and the other, with a mass of $80 \mathrm{kg},\( is approaching her with a speed of \)2.0 \mathrm{m} / \mathrm{s}$ at an angle of \(20^{\circ}\) above the horizontal. What is the direction and speed of the acrobats right after they grab on to each other? Let the positive \(x\) -axis be in the horizontal direction and assume the first acrobat has positive velocity components in the positive \(x\)- and \(y\)-directions.
A 60.0 -kg woman stands at one end of a \(120-\) kg raft that is $6.0 \mathrm{m}\( long. The other end of the raft is \)0.50 \mathrm{m}$ from a pier. (a) The woman walks toward the pier until she gets to the other end of the raft and stops there. Now what is the distance between the raft and the pier? (b) In (a), how far did the woman walk (relative to the pier)?
A block of wood of mass \(0.95 \mathrm{kg}\) is initially at rest. A bullet of mass \(0.050 \mathrm{kg}\) traveling at \(100.0 \mathrm{m} / \mathrm{s}\) strikes the block and becomes embedded in it. With what speed do the block of wood and the bullet move just after the collision?
See all solutions

Recommended explanations on Physics Textbooks

Further Mechanics and Thermal Physics

Read Explanation

Medical Physics

Read Explanation

Turning Points in Physics

Read Explanation

Translational Dynamics

Read Explanation

Energy Physics

Read Explanation

Force

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.