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Chapter 13: Problem 52

A \(1.80-\mathrm{kg}\) monkey wrench is pivoted 0.250 \(\mathrm{m}\) from its center of mass and allowed to swing as a physical pendulum. The period for small-angle oscillations is 0.940 s. (a) What is the moment of inertia of the wrench about an axis through the pivot? (b) If the wrench is initially displaced 0.400 rad from its equilibrium position, what is the angular speed of the wrench as it passes through the equilibrium position?

Short Answer

Expert verified
(a) Moment of inertia \(I \approx 0.015\,\text{kg} \cdot \text{m}^2\). (b) Angular speed \(\omega \approx 2.45\,\text{rad/s}\).

Step by step solution

01

Understanding the Given Information

We are given a monkey wrench with a mass of \(1.80\,\text{kg}\), pivoted at \(0.250\,\text{m}\) from its center of mass. The period of oscillation for a small-angle pendulum is \(0.940\,\text{s}\).
02

Relate Period to Moment of Inertia

The period \(T\) of a physical pendulum is given by the formula:\[T = 2\pi \sqrt{\frac{I}{mgh}}\]where \(I\) is the moment of inertia about the pivot point, \(m\) is the mass, \(g\) is the acceleration due to gravity (\(9.81\,\text{m/s}^2\)), and \(h\) is the distance from the pivot to the center of mass (\(0.250\,\text{m}\)).
03

Solve for Moment of Inertia

Rearrange the period formula to solve for \(I\):\[I = \frac{(T^2 \cdot m \cdot g \cdot h)}{4\pi^2}\]Substitute \(T = 0.940\,\text{s}\), \(m = 1.80\,\text{kg}\), \(g = 9.81\,\text{m/s}^2\), and \(h = 0.250\,\text{m}\):\[I = \frac{(0.940^2 \cdot 1.80 \cdot 9.81 \cdot 0.250)}{4\pi^2}\]Calculate \(I\).
04

Initial Displacement and Angular Speed

For small oscillations, when the wrench is initially displaced by \(\theta = 0.400\,\text{rad}\), energy conservation principles can be applied. At maximum displacement, all energy is potential: \[U = mgh(1 - \cos\theta)\]As it passes through equilibrium, all energy is kinetic:\[K = \frac{1}{2} I \omega^2\]Setting \(U = K\) and solving for \(\omega\):
05

Calculate Angular Speed

Equating potential and kinetic energy and solving for \(\omega\):\[mgh(1 - \cos\theta) = \frac{1}{2} I \omega^2\]Solve:\[\omega = \sqrt{\frac{2mgh(1 - \cos\theta)}{I}}\]Use the calculated \(I\) from Step 3 and substitute the values \(h = 0.250\,\text{m}\), \(m = 1.80\,\text{kg}\), and \(\theta = 0.400\,\text{rad}\) into the equation to find \(\omega\).
06

Final Calculation and Solution

Calculate the numerical value of the moment of inertia \(I\) using the equation from Step 3. Then, use the value of \(I\) to find angular speed \(\omega\) at equilibrium using the formula from Step 5. The values will be:- (a) Moment of Inertia \(I\)- (b) Angular speed \(\omega\)

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Key Concepts

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

Moment of Inertia
In physics, the moment of inertia is a key concept when dealing with rotating objects, such as a pendulum. It tells us how much torque, or rotational force, is required to make the object rotate at a certain rate. For a physical pendulum like the monkey wrench mentioned in the problem, its moment of inertia depends on its mass and how that mass is distributed relative to the pivot point.
  • It is denoted by the letter \( I \).
  • Its formula in this context is derived from the period of the pendulum: \[ I = \frac{(T^2 \cdot m \cdot g \cdot h)}{4\pi^2} \]
  • In the problem, the parameters included are a period \( T \), a mass \( m \), gravitational acceleration \( g \), and distance from pivot \( h \).
Knowing \( I \) helps in understanding and predicting the motion of the pendulum.
Oscillation Period
The oscillation period refers to the time it takes for one complete cycle of motion. In our context, it's the time for the monkey wrench to swing from one side, back to its starting point. The period \( T \) of a physical pendulum depends on several factors:
  • Moment of Inertia \( I \)
  • Mass \( m \) of the pendulum
  • Distance from the pivot to the center of mass \( h \)
  • Gravitational acceleration \( g \)
The formula \( T = 2\pi \sqrt{\frac{I}{mgh}} \) considers these components. Understanding this period is essential for predicting how swiftly or slowly the pendulum swings. A larger moment of inertia or greater mass would increase the period, causing the pendulum to swing more slowly.
Angular Displacement
Angular displacement in pendulums refers to how much the pendulum swings away from its equilibrium position. It is measured in radians. At its initial position, the monkey wrench is displaced by \( 0.400 \) rad, as per the exercise.When considering angular displacement:
  • It tells us how *far* the pendulum swings, which is crucial for calculating its energy potential.
  • Describes the initial angle of swing relative to the vertical.
  • High displacement means greater potential energy initially.
Understanding angular displacement helps us calculate energy conservation and predict how fast the pendulum will move when passing through the equilibrium position later on.
Energy Conservation
Energy conservation is a fundamental principle in physics that applies neatly to pendulums. It states that energy in a closed system remains the same, transferring between forms but never being destroyed.For the pendulum:
  • Initially, all energy is potential when displaced at an angle.
  • When crossing equilibrium, this transforms to kinetic energy.
The principle can be shown by the equation:\[ mgh(1 - \cos\theta) = \frac{1}{2} I \omega^2 \]where \( \omega \) is the angular speed. This relation helps in calculating how fast the wrench will be moving as it swings past its lowest point. By setting potential energy equal to kinetic energy, we can solve for angular velocity.
Small-Angle Approximation
In pendulum problems, the small-angle approximation simplifies mathematical calculations, making it manageable to analyze pendulum behavior. This approximation assumes that when angles are small (usually less than 15掳 or 0.26 rad), the sine of the angle is approximately equal to the angle itself when measured in radians:\( \sin\theta \approx \theta \)Applying this here:
  • Allows simple harmonic motion assumptions, making solutions like the oscillation period accurate.
  • Simplifies complex differential equations to easier forms.
By using the small-angle approximation, we streamline otherwise complex calculations into simpler equations while still delivering close approximations of the actual physical behavior for pendulums.

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Most popular questions from this chapter

A sinusoidally varying driving force is applied to a damped harmonic oscillator of force constant \(k\) and mass \(m\) . If the damping constant has a value \(b_{1}\) , the amplitude is \(A_{1}\) when the driving angular frequency equals \(\sqrt{k / m} .\) In terms of \(A_{1},\) what is the amplitude for the same driving frequency and the same driving force amplitude \(F_{\text { max }}\) if the damping constant is \((a) 3 b_{1}\) and \((b) b_{1} / 2 ?\)

The tip of a tuning fork goes through 440 complete vibra- tions in 0.500 s. Find the angular frequency and the period of the motion.

A child with poor table manners is sliding his 250 -g dinner plate back and forth in SHM with an amplitude of 0.100 \(\mathrm{m}\) on a horizontal surface. At a point 0.060 \(\mathrm{m}\) away from equilibrium, the speed of the plate is 0.300 \(\mathrm{m} / \mathrm{s}\) . (a) What is the period? (b) What is the displacement when the speed is 0.160 \(\mathrm{m} / \mathrm{s} ?(\mathrm{c})\) In the center of the dinner plate is a \(10.0-\mathrm{g}\) carrot slice. If the carrot slice is just on the verge of slipping at the endpoint of the path, what is the coefficient of static friction between the carrot slice and the plate?

A cheerleader waves her pom-pom in SHM with an amplitude of 18.0 \(\mathrm{cm}\) and a frequency of 0.850 \(\mathrm{Hz}\) . Find (a) the maximum magnitude of the acceleration and of the velocity; \((6)\) the acceleration and speed when the poin-poin's coordinate is \(x=+9.0 \mathrm{cm}\); (c) the time required to move from the equilibrium position directly to a point 12.0 \(\mathrm{cm}\) away. (d) Which of the quantities asproach used parts \((\mathrm{a}),(\mathrm{b}),\) and \((\mathrm{c})\) can be found using the energy approach used in Section \(13.3,\) and which cannot? Explain.

The two pendulums shown in Fig 13.34 each consist of a uniform solid ball of mass \(M\) supported by a massless string, but the ball for pendulum \(A\) is very tiny while the ball for pendulum \(B\) is much larger. Find the period of each pendulum for small displacements. Which ball takes longer to complete a swing?
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