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Chapter 14: Problem 58

A holiday ornament in the shape of a hollow sphere with mass \(M =\) 0.015 kg and radius \(R =\) 0.050 m is hung from a tree limb by a small loop of wire attached to the surface of the sphere. If the ornament is displaced a small distance and released, it swings back and forth as a physical pendulum with negligible friction. Calculate its period. (\(Hint\): Use the parallel-axis theorem to find the moment of inertia of the sphere about the pivot at the tree limb.)

Short Answer

Expert verified
The period of the pendulum is approximately 0.451 seconds.

Step by step solution

01

Understanding the Physical Pendulum

A physical pendulum is an object that can oscillate about an axis. The period of a physical pendulum depends on its moment of inertia about the pivot point and the distance from the pivot to its center of mass. The formula for the period is given by: \( T = 2\pi \sqrt{\frac{I}{mgd}} \), where \(I\) is the moment of inertia, \(m\) is the mass, \(g\) is the acceleration due to gravity (9.81 m/s虏), and \(d\) is the distance from the pivot to the center of mass.
02

Moment of Inertia via the Parallel-Axis Theorem

The moment of inertia for a solid sphere about its center is \( I_{cm} = \frac{2}{5}MR^2 \). Using the parallel-axis theorem, we find the moment of inertia about the pivot point at the surface of the sphere: \( I = I_{cm} + Md^2 \). Since the sphere is hollow, we assume it's thin, contributing mostly at the radius, thus \( d = R \). Therefore, \( I = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2 \).
03

Calculate the Moment of Inertia

Substitute the values into the moment of inertia formula: \( I = \frac{7}{5}(0.015) \times (0.050)^2 \). This calculation gives us \( I \approx 1.05 \times 10^{-5} \text{ kg m}^2 \).
04

Calculate the Period of the Pendulum

Substitute \( I = 1.05 \times 10^{-5} \), \( m = 0.015 \text{ kg} \), \( g = 9.81 \text{ m/s}^2 \), and \( d = 0.050 \text{ m} \) into the period equation: \( T = 2\pi \sqrt{\frac{1.05 \times 10^{-5}}{0.015 \times 9.81 \times 0.050}} \). This simplifies to \( T \approx 0.451 \text{ seconds} \).

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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 measure of how difficult it is to change the rotation of an object. It's often thought as a rotational analogue to mass in linear motion. For different shapes and axis configurations, the moment of inertia can vary greatly.

For a sphere, this quantity is central in the calculations of a physical pendulum. The moment of inertia depends on both the mass distribution of the object and the axis about which it rotates.
  • The base formula for a solid sphere rotating about its own center is: \( I_{cm} = \frac{2}{5}MR^2 \).
  • For our physical pendulum, because the rotation isn't around the center, we need to adjust using the parallel-axis theorem.

This adapted formula enables correct calculations for rotations different from the base axis.
Parallel-Axis Theorem
The parallel-axis theorem is a powerful tool in physics that allows for calculating the moment of inertia of an object when it is rotating about an axis that is parallel but not through the center of mass.

The formula is: \[ I = I_{cm} + Md^2 \] where:
  • \(I_{cm}\) is the moment of inertia through the center of mass,
  • \(M\) is the mass of the object, and
  • \(d\) is the distance from the new axis to the axis through the center of mass.

In the given problem, the sphere is hung and rotates around an axis through the hanging loop, which is at its surface; hence the distance \(d\) equals \(R\). Applying the parallel-axis theorem simplifies the calculation, providing the precise moment of inertia needed for the subsequent steps.
Calculating Period
To determine how long it takes for the pendulum to complete one full swing back and forth, we calculate its period. This involves using key parameters derived from its physical properties.
The formula for a physical pendulum is: \[ T = 2\pi \sqrt{\frac{I}{mgd}} \] where:
  • \(T\) is the period,
  • \(I\) is the moment of inertia,
  • \(m\) is mass,
  • \(g\) is the acceleration due to gravity, and
  • \(d\) is the distance from the pivot to the center of mass.

This equation perfectly encapsulates how both the distribution of mass and the path it swings through impact the pendulum's movement. With the values substituted from the physical attributes of the sphere, we derive the period as approximately 0.451 seconds鈥攁 swift oscillation due to its small size and mass.
Physics Problem Solving
Solving physics problems often requires a mix of theoretical knowledge and practical grasp of concepts. This problem specifically illustrates applying several principles in a sequence to find a direct answer, demonstrating the blend of understanding and calculation work required in physics.

Here's a breakdown of the approach:
  • Comprehend the problem context, identifying it as a physical pendulum scenario.
  • Calculate the fundamental parameters like the moment of inertia using relevant theorems.
  • Integrate these parameters into broader formulas, like the one for period calculation.
Understanding these steps and how they connect fosters a deeper comprehension of wider physics concepts, such as oscillations and rotational dynamics, preparing students for tackling more complex topics.

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Most popular questions from this chapter

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

A 0.500-kg mass on a spring has velocity as a function of time given by \({v_x}(t) = -\)(3.60 cm/s) sin[ (4.71 rad/s)\(t - \pi\)/2) ]. What are (a) the period; (b) the amplitude; (c) the maximum acceleration of the mass; (d) the force constant of the spring?

An unhappy 0.300 -kg rodent, moving on the end of a spring with force constant \(k=2.50 \mathrm{~N} / \mathrm{m},\) is acted on by a damping force \(F_{x}=-b v_{x}\). (a) If the constant \(b\) has the value \(0.900 \mathrm{~kg} / \mathrm{s}\), what is the frequency of oscillation of the rodent? (b) For what value of the constant \(b\) will the motion be critically damped?

In February 2004, scientists at Purdue University used a highly sensitive technique to measure the mass of a vaccinia virus (the kind used in smallpox vaccine). The procedure involved measuring the frequency of oscillation of a tiny sliver of silicon (just 30 nm long) with a laser, first without the virus and then after the virus had attached itself to the silicon. The difference in mass caused a change in the frequency. We can model such a process as a mass on a spring. (a) Show that the ratio of the frequency with the virus attached \((f_S+\,_V)\) to the frequency without the virus \((f_S)\) is given by \(f_S+\,_V/f_S = 1\sqrt{ 1 + (m_V/m_S) }\), where \(m_V\) is the mass of the virus and \(m_S\) is the mass of the silicon sliver. Notice that it is \(not\) necessary to know or measure the force constant of the spring. (b) In some data, the silicon sliver has a mass of 2.10 \(\times\) 10\(^{-16}\) g and a frequency of 2.00 \(\times\) 10\(^{15}\) Hz without the virus and 2.87 \(\times\) 10\(^{14}\) Hz with the virus. What is the mass of the virus, in grams and in femtograms?

A mass is oscillating with amplitude \(A\) at the end of a spring. How far (in terms of \(A\)) is this mass from the equilibrium position of the spring when the elastic potential energy equals the kinetic energy?
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