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Chapter 14: Problem 33

Interestingly, there have been several studies using cadavers to determine the moment of inertia of human body parts by letting them swing as a pendulum about a joint. In one study, the center of gravity of a \(5.0 \mathrm{kg}\) lower leg was found to be \(18 \mathrm{cm}\) from the knee. When pivoted at the knee and allowed to swing, the oscillation frequency was \(1.6 \mathrm{Hz}\). What was the moment of inertia of the lower leg?

Short Answer

Expert verified
The moment of inertia of the lower leg is approximately \(0.145 \, kg.m虏\).

Step by step solution

01

Formulate the equation

The moment of inertia \( I \) of a pendulum is given by the formula \( I = 4蟺虏mf虏l/g \) where \( m \) is mass, \( f \) is frequency, \( l \) is length and \( g \) is the acceleration due to gravity. The values are given as \( m = 5.0 \, kg \), \( f = 1.6 \, Hz \), \( l = 18 \, cm = 0.18 \, m \) (converted into meters being SI unit of length) and \( g = 9.81 \, m/s虏 \).
02

Substitute the values

Substitute these values into the equation, it gives \( I = 4蟺虏(5)(1.6)^2(0.18)/9.81 \).
03

Calculate the moment of inertia

Using a calculator to solve the equation gives the moment of inertia of the lower leg.

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Key Concepts

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

Pendulum Motion
Pendulum motion describes the back and forth swing of an object suspended from a fixed point. Imagine a simple pendulum as a weight suspended from a string or rod, allowed to move back and forth under the influence of gravity. When released from a certain angle, the pendulum swings, creating an oscillatory motion. The motion can be consistent, creating a rhythmic pattern. This simple movement can be applied to many physical scenarios, such as the swinging of a human limb as studied with cadavers. The study of pendulum motion not only aids in understanding simple harmonic motion but also helps in identifying physical properties, such as the moment of inertia in our exercise.
Oscillation Frequency
In pendulum motion, oscillation frequency represents how often the pendulum swings back and forth in a given time frame. It's the number of complete oscillations per second, measured in Hertz (Hz). Frequency is vital in the study of pendulum properties. It helps determine other physical attributes. In our exercise, the frequency of the limb's swing is found to be 1.6 Hz. The frequency influences the calculation of the moment of inertia. By knowing the oscillation frequency, alongside other values (mass and distance from pivot), we can compute the moment of inertia. It shows how frequency forms a crucial part of pendulum-related physics problems.
Center of Gravity
The center of gravity is the point where the weight of an object is considered to be concentrated. For pendular motion and physics in general, understanding this concept is crucial. It is the balance point. In our given scenario, the center of gravity is crucial in determining how the lower leg pivots around the knee. The leg's center of gravity is located 18 cm from the knee. Identifying this point allows for accurate calculation of how a body will swing, affecting both the oscillation frequency and the moment of inertia. The center of gravity plays a pivotal role in understanding the balance and dynamics of physical objects.
Physics Formulas
Physics formulas are concise tools used to express relationships between different physical quantities. They allow us to calculate unknown variables if we have enough known quantities. In this exercise, the formula for the moment of inertia (\( I = \frac{4\pi^2 m f^2 l}{g} \)) is used. By plugging in the mass, oscillation frequency, and length, it simplifies the calculation. Converting all measurements to the same unit system (like meters for length in SI units) is important. Physics formulas can seem complex but provide a systematic approach to solving real-world physics problems. Understanding how to apply these formulas is key for finding precise solutions.

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Most popular questions from this chapter

The position of a 50 g oscillating mass is given by \(x(t)=\) \((2.0 \mathrm{cm}) \cos (10 t),\) where \(t\) is in seconds. Determine: a. The amplitude. b. The period. c. The spring constant. d. The maximum speed. e. The total energy. f. The velocity at \(t=0.40 \mathrm{s}\)

The ultrasonic transducer used in a medical ultrasound imaging device is a very thin disk \((m=0.10 \mathrm{g})\) driven back and forth in SHM at 1.0 MHz by an electromagnetic coil. a. The maximum restoring force that can be applied to the disk without breaking it is \(40,000 \mathrm{N}\). What is the maximum oscillation amplitude that won't rupture the disk? b. What is the disk's maximum speed at this amplitude?

A \(1.0 \mathrm{kg}\) block is attached to a spring with spring constant \(16 \mathrm{N} / \mathrm{m} .\) While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of \(40 \mathrm{cm} / \mathrm{s}\). What are a. The amplitude of the subsequent oscillations? b. The block's speed at the point where \(x=\frac{1}{2} A ?\)

A block with a mass of \(0.28 \mathrm{kg}\) is attached to a horizontal spring. The block is pulled back from its equilibrium position until the spring exerts a force of \(1.0 \mathrm{N}\) on the block. When the block is released, it oscillates with a frequency of \(1.2 \mathrm{Hz} .\) How far was the block pulled back before being released?

A \(200 \mathrm{g}\) mass attached to a horizontal spring oscillates at a frequency of \(2.0 \mathrm{Hz}\). At one instant, the mass is at \(x=5.0 \mathrm{cm}\) and has \(v_{x}=-30 \mathrm{cm} / \mathrm{s} .\) Determine: a. The period. b. The amplitude. c. The maximum speed. d. The total energy.
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