/*! 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 50 . Changing your center of mass. ... [FREE SOLUTION] | 91Ó°ÊÓ

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. Changing your center of mass. To keep the calculations fairly simple, but still reasonable, we shall model a human leg that is 92.0 \(\mathrm{cm}\) long (measured from the hip joint) by assuming that he upper leg and the lower leg (which includes the foot) have equal lengths and that each of them is uniform. For a 70.0 kg per- son, the mass of the upper leg would be 8.60 \(\mathrm{kg}\) , while that of the lower leg (including the foot) would be 5.25 \(\mathrm{kg}\) . Find the location of the center of mass of this leg, relative to the hip joint, if it is (a) stretched out horizontally and (b) bent at the knee to form a right angle with the upper leg remaining horizontal.

Short Answer

Expert verified
(a) 40.5 cm, (b) 31.7 cm from the hip joint.

Step by step solution

01

Understand the problem

We are tasked to find the center of mass of a leg modeled as two uniform segments: the upper leg and the lower leg, which includes the foot. Each segment is of equal length, 46.0 cm, as they are halves of the total leg length (92.0 cm). The upper leg weighs 8.60 kg, and the lower leg weighs 5.25 kg. We need to calculate the center of mass in two scenarios: (a) the leg is stretched out horizontally, and (b) the leg is bent at a right angle at the knee.
02

Define known values and segment lengths

- Length of each segment: 46.0 cm- Mass of the upper leg, \(m_1 = 8.60\, \text{kg}\)- Mass of the lower leg, \(m_2 = 5.25\, \text{kg}\)For each scenario, we'll calculate the position of the center of mass using the formula:\[\bar{x} = \frac{m_1 \cdot x_1 + m_2 \cdot x_2}{m_1 + m_2}\]
03

Calculate center of mass for horizontal leg (stretched out)

In scenario (a), both segments are horizontal. Therefore, the position of their centers of mass from the hip joint are:- Upper leg, \(x_1 = 23.0\, \text{cm}\) (midpoint of 46.0 cm)- Lower leg, \(x_2 = 69.0\, \text{cm}\) (23.0 cm + 46.0 cm)Substitute into the formula:\[\bar{x} = \frac{8.60 \cdot 23.0 + 5.25 \cdot 69.0}{8.60 + 5.25} = \frac{197.8 + 362.25}{13.85} \approx 40.5 \, \text{cm}\]
04

Calculate center of mass for bent leg at right angle

In scenario (b), we need to account for the leg being bent. The upper leg remains horizontal, so:- \(x_1 = 23.0\, \text{cm}\)For the lower leg, it's bent down 90°, meaning its center of mass is vertically aligned with its bending end, thus:- \(x_2 = 46.0\, \text{cm}\) horizontally (full length position relative to hip)Substitute these values into the formula:\[\bar{x} = \frac{8.60 \cdot 23.0 + 5.25 \cdot 46.0}{8.60 + 5.25} = \frac{197.8 + 241.5}{13.85} \approx 31.7 \, \text{cm}\]
05

Interpret the results

For scenario (a), the center of mass is located at 40.5 cm from the hip when the leg is horizontally stretched out. In scenario (b), the center of mass shifts closer to the hip to 31.7 cm when the leg is bent at the knee in a right angle. This shift happens because the lower leg's mass center is positioned directly below the knee, reducing its horizontal effective position.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Human Body Mechanics
Understanding human body mechanics is essential to analyzing how we move and interact with our environment. This field studies how different parts of the human body, like bones and muscles, work together to support movement. In the context of the given exercise, mechanics involve analyzing how the leg is structured and moves at the hip and knee joints.
The leg can be thought of as a system of levers, where bones act as rigid bars, joints as fulcrums, and muscles provide the force to move these levers. The hip and knee serve as primary joints for leg movement, allowing both horizontal stretches and bends. Understanding how the leg's mechanics work helps determine how forces are distributed across the body during different movements.
Biomechanics
Biomechanics combines the study of biology with the principles of mechanics to understand movement better. It aims to explain how forces within the body influence how we move and interact with external physical forces. This framework is crucial for studying the leg movement described in the exercise.
In biomechanics, we often simplify and generalize body parts to model them for calculations. For instance, dividing the leg into two segments helps us understand how its mass is distributed and moves relative to the body during different activities. By breaking down complex structures into manageable parts, we can predict and analyze the body's mechanical function more accurately. This helps in fields like sports science and medical rehabilitation.
Mass Distribution
Mass distribution is about how the mass of an object is spread throughout its structure, affecting its balance and stability. In human anatomy, how mass is distributed across body parts can significantly impact movement and balance.
For the exercise at hand, understanding the mass distribution of the upper and lower leg allows us to calculate the center of mass. The center of mass is a crucial point where the mass of the body is considered to be concentrated. Calculating this for the leg involves finding a balance point where the mass of the upper and lower segments is equal. This balance point shifts when the leg bends, illustrating how dynamic and responsive our body is to different postures and movements.
Calculations in Physics
Physics provides the tools needed to model and solve problems related to movement and forces effectively. In this exercise, using physics formulas allows us to calculate the center of mass of the human leg in different positions.
To find the center of mass, the formula \[\bar{x} = \frac{m_1 \cdot x_1 + m_2 \cdot x_2}{m_1 + m_2}\] helps combine the weighted positions of each segment. Here, \(m_1\) and \(m_2\) represent the masses of the upper and lower leg, while \(x_1\) and \(x_2\) denote their respective positions.
This calculation simplifies the complex interactions in our body to manageable equations, making it easier to predict how our body behaves in different poses or movements. These fundamental physics concepts are vital in studying how we move and designing systems, like prosthetics, that emulate natural human movement.

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Most popular questions from this chapter

A rifle bullet with mass 8.00 g strikes and embeds itself in a block with a mass of 0.992 kg that rests on a frictionless, horizontal surface and is attached to a coil spring. (See Figure \(8.47 . )\) The impact compresses the spring 15.0 \(\mathrm{cm} .\) Calibration of the spring shows that a force of 0.750 \(\mathrm{N}\) is required to compress the spring 0.250 \(\mathrm{cm}\) . (a) Find the magnitude of the block's velocity just after impact. (b) What was the initial speed of the bullet?

\(\cdot\) Three identical boxcars are coupled together and are moving at a constant speed of 20.0 \(\mathrm{m} / \mathrm{s}\) on a level track. They collide with another identical boxcar that is initially at rest and couple to it, so that the four cars roll on as a unit. Friction is small enough to be neglected. (a) What is the speed of the four cars? (b) What percentage of the kinetic energy of the boxcars is dissipated in the collision? What happened to this energy?

A block of ice with a mass of 2.50 \(\mathrm{kg}\) is moving on a fric- tionless, horizontal surface. At \(t=0,\) the block is moving to the right with a velocity of magnitude 8.00 \(\mathrm{m} / \mathrm{s}\) . Calculate the magnitude and direction of the velocity of the block after each of the following forces has been applied for 5.00 s: (a) a force of 5.00 N directed to the right; (b) a force of 7.00 \(\mathrm{N}\) directed to the left.

Combining conservation laws. A 15.0 kg block is attached to a very light horizontal spring of force constant 500.0 \(\mathrm{N} / \mathrm{m}\) and is resting on a frictionless horizontal table. (See Figure \(8.38 .\) ) Suddenly it is struck by a 3.00 \(\mathrm{kg}\) stone traveling horizontally at 8.00 \(\mathrm{m} / \mathrm{s}\) to the right, whereupon the stone rebounds at 2.00 \(\mathrm{m} / \mathrm{s}\) horizontally to the left. Find the maximum distance that the block will compress the spring after the collision. (Hint: Break this problem into two parts the collision and the behavior after the collision \(-\) and apply the appropriate conservation law to each part.)

. Biomechanics. The mass of a regulation tennis ball is 57 g (although it can vary slightly), and tests have shown that the ball is in contact with the tennis racket for 30 ms. (This number can also vary, depending on the racket and swing.) We shall assume a 30.0 \(\mathrm{ms}\) contact time throughout this problem. The fastest-known served tennis ball was served by "Big Bill" Tilden in \(1931,\) and its speed was measured to be 73.14 \(\mathrm{m} / \mathrm{s}\) .(a) What impulse and what force did Big Bill exert on the ten- nis ball in his record serve? (b) If Big Bill's opponent returned his serve with a speed of \(55 \mathrm{m} / \mathrm{s},\) what force and what impulse did he exert on the ball, assuming only horizontal motion?

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