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Ch 15: Oscillations
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 15, Problem 57

Interestingly, there have been several studies using cadavers to determine the moments of inertia of human body parts, information that is important in biomechanics. In one study, the center of mass of a 5.0 kg lower leg was found to be 18 cm from the knee. When the leg was allowed to pivot at the knee and swing freely as a pendulum, the oscillation frequency was 1.6 Hz. What was the moment of inertia of the lower leg about the knee joint?

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Step 1: Understand the problem. The goal is to find the moment of inertia (I) of the lower leg about the knee joint. The given data includes the mass of the leg (m = 5.0 kg), the distance from the knee to the center of mass (d = 18 cm = 0.18 m), and the oscillation frequency (f = 1.6 Hz). The leg is modeled as a physical pendulum.
Step 2: Recall the relationship between the oscillation frequency and the moment of inertia for a physical pendulum. The formula is: \( f = \frac{1}{2\pi} \sqrt{\frac{m g d}{I}} \), where \( g \) is the acceleration due to gravity (approximately 9.8 m/s²), \( m \) is the mass, \( d \) is the distance to the center of mass, and \( I \) is the moment of inertia.
Step 3: Rearrange the formula to solve for the moment of inertia \( I \). Start by isolating \( I \): \( I = \frac{m g d}{(2\pi f)^2} \). This equation will allow us to calculate the moment of inertia once all values are substituted.
Step 4: Substitute the known values into the formula. Use \( m = 5.0 \ \text{kg} \), \( g = 9.8 \ \text{m/s}^2 \), \( d = 0.18 \ \text{m} \), and \( f = 1.6 \ \text{Hz} \). The denominator \( (2\pi f)^2 \) will involve squaring \( 2\pi \times 1.6 \).
Step 5: Perform the calculations step by step to find \( I \). First, calculate \( 2\pi f \), then square it. Next, calculate \( m g d \). Finally, divide \( m g d \) by \( (2\pi f)^2 \) to find the moment of inertia. Ensure all units are consistent throughout the calculation.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Moment of Inertia

Moment of inertia is a measure of an object's resistance to changes in its rotational motion about an axis. It depends on the mass distribution relative to the axis of rotation; the further the mass is from the axis, the greater the moment of inertia. In biomechanics, calculating the moment of inertia is crucial for understanding how body parts behave during movement.
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Intro to Moment of Inertia

Center of Mass

The center of mass is the point at which the mass of an object is concentrated and around which its mass is evenly distributed. For a human limb, like the lower leg, the center of mass affects how it swings and rotates. Knowing the center of mass is essential for calculating the moment of inertia and analyzing motion dynamics.
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Pendulum Motion

Pendulum motion refers to the oscillatory movement of an object swinging back and forth around a pivot point. The frequency of oscillation is influenced by the length of the pendulum and the acceleration due to gravity. In this context, the frequency of the lower leg's swing can be used to derive the moment of inertia using the principles of rotational dynamics.
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Related Practice
Textbook Question

A mass hanging from a spring oscillates with a period of 0.35 s. Suppose the mass and spring are swung in a horizontal circle, with the free end of the spring at the pivot. What rotation frequency, in rpm, will cause the spring's length to stretch by 15%?

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Textbook Question

A uniform rod of mass M and length L swings as a pendulum on a pivot at distance L/4 from one end of the rod. Find an expression for the frequency f of small-angle oscillations.

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Textbook Question

A 500 g air-track glider attached to a spring with spring constant 10 N/m is sitting at rest on a frictionless air track. A 250 g glider is pushed toward it from the far end of the track at a speed of 120 cm/s. It collides with and sticks to the 500 g glider. What are the amplitude and period of the subsequent oscillations?

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Textbook Question

It has recently become possible to 'weigh' DNA molecules by measuring the influence of their mass on a nano-oscillator. FIGURE P15.58 shows a thin rectangular cantilever etched out of silicon (density 2300 kg/m³) with a small gold dot (not visible) at the end. If pulled down and released, the end of the cantilever vibrates with SHM, moving up and down like a diving board after a jump. When bathed with DNA molecules whose ends have been modified to bind with gold, one or more molecules may attach to the gold dot. The addition of their mass causes a very slight—but measurable—decrease in the oscillation frequency. A vibrating cantilever of mass M can be modeled as a block of mass ⅓M attached to a spring. (The factor of ⅓ arises from the moment of inertia of a bar pivoted at one end.) Neither the mass nor the spring constant can be determined very accurately—perhaps to only two significant figures—but the oscillation frequency can be measured with very high precision simply by counting the oscillations. In one experiment, the cantilever was initially vibrating at exactly 12 MHz. Attachment of a DNA molecule caused the frequency to decrease by 50 Hz. What was the mass of the DNA?

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Textbook Question

A 1.00 kg block is attached to a horizontal spring with spring constant 2500 N/m. The block is at rest on a frictionless surface. A 10 g bullet is fired into the block, in the face opposite the spring, and sticks. What was the bullet's speed if the subsequent oscillations have an amplitude of 10.0 cm?

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Textbook Question

A compact car has a mass of 1200 kg. Assume that the car has one spring on each wheel, that the springs are identical, and that the mass is equally distributed over the four springs. What will be the car's oscillation frequency while carrying four 70 kg passengers?

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