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A holiday ornament in the shape of a hollow sphere with mass \(M=0.015 \mathrm{kg}\) and radius \(R=0.050 \mathrm{m}\) 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 reased, 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 ornament's swing is approximately 1.14 seconds.

Step by step solution

01

Understand the Problem

We need to find the period of a hollow sphere functioning as a physical pendulum. The mass of the sphere is given as \( M = 0.015 \mathrm{kg} \) and its radius is \( R = 0.050 \mathrm{m} \). The solution involves finding the moment of inertia about the pivot point using the parallel-axis theorem.
02

Moment of Inertia of a Hollow Sphere

The moment of inertia of a hollow sphere about its own center is \[ I_c = \frac{2}{3}MR^2. \]Given that \( M = 0.015 \mathrm{kg} \) and \( R = 0.050 \mathrm{m} \), we substitute these values to calculate \( I_c \).
03

Applying Parallel-Axis Theorem

The parallel-axis theorem states that the moment of inertia about a parallel axis at distance \( d \) is \[ I = I_c + Md^2. \]Here, \( d = R \), so \[ I = \frac{2}{3}MR^2 + MR^2 = \frac{5}{3}MR^2. \]
04

Calculate the Moment of Inertia about the Pivot

Substituting the mass \( M = 0.015 \mathrm{kg} \) and radius \( R = 0.050 \mathrm{m} \) into the equation from Step 3, we find \[ I = \frac{5}{3}(0.015)(0.050)^2 = 6.25 \times 10^{-5} \mathrm{kg} \cdot \mathrm{m}^2. \]
05

Use the Formula for the Period of a Physical Pendulum

The period \( T \) of a physical pendulum is given by \[ T = 2\pi \sqrt{\frac{I}{MgL}}, \]where \( I \) is the moment of inertia about the pivot, \( M \) is the mass, \( g \) is the acceleration due to gravity (approximately \(9.81 \mathrm{m/s^2}\)), and \( L = R = 0.050 \mathrm{m} \) is the distance from the pivot to the center of mass.
06

Calculate the Period

Substitute \( I = 6.25 \times 10^{-5} \mathrm{kg} \cdot \mathrm{m}^2 \), \( M = 0.015 \mathrm{kg} \), \( g = 9.81 \mathrm{m/s^2} \), and \( L = 0.050 \mathrm{m} \) into the equation for \( T \):\[ T = 2\pi \sqrt{\frac{6.25 \times 10^{-5}}{0.015 \cdot 9.81 \cdot 0.050}}. \] Calculate to find that \( T \approx 1.14 \mathrm{s} \).

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Key Concepts

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

Moment of Inertia
When we talk about the moment of inertia, we're focusing on how an object's mass is distributed in relation to a specific axis. This is crucial in the realm of physics as it impacts the object's rotational movement. For a hollow sphere, which is essentially an empty ball like the holiday ornament mentioned in the exercise, the moment of inertia about its center is defined by the formula:\[ I_c = \frac{2}{3}MR^2. \]Here, the letter "I" represents the moment of inertia, "M" is the mass of the sphere, and "R" stands for the radius. In simple terms, it measures the object's resistance to changes in its rotational motion. Understanding the moment of inertia provides insights into how the hollow sphere behaves when it swings, and is essential for calculating its period as a physical pendulum.
Parallel-Axis Theorem
The parallel-axis theorem is a mathematical principle that helps us find the moment of inertia of an object about an axis that is parallel to one through the center of mass. This is particularly useful in many physics problems, including the one involving our holiday ornament.This theorem states:\[ I = I_c + Md^2. \]Where:- \( I \) is the moment of inertia about the new axis.- \( I_c \) is the moment of inertia about the object's center of mass.- \( M \) is the mass.- \( d \) is the distance between the two axes.In the case of the hollow sphere, the pivot axis is a distance \( R \) from the sphere's center. Applying the theorem helps adjust the sphere's inertia to this new axis, transforming it from center-based to pivot-based, which is necessary for further calculations in pendulum dynamics.
Period of a Pendulum
When we want to find out how long it takes for a pendulum to complete one full swing back and forth, we're interested in its period. For a physical pendulum like our hollow sphere, this period is determined by more than just its length and how high it's hung.The formula for calculating the period \( T \) is:\[ T = 2\pi \sqrt{\frac{I}{MgL}}, \]where:- \( I \) is the moment of inertia about the pivot point.- \( M \) is the mass of the pendulum.- \( g \) is the acceleration due to gravity, approximately \( 9.81 \mathrm{m/s^2} \).- \( L \) is the distance from the pivot to the center of mass.By knowing these factors, we can accurately determine how quickly or slowly the ornament swings. This understanding is vital in cases where precise timing of the pendulum motion is required, such as in timekeeping devices or understanding simple harmonic motion behavior.
Hollow Sphere
A hollow sphere, like our Christmas ornament, has a unique physical structure: all its mass is concentrated on its surface, while the interior is empty. This contrasts with a solid sphere, where mass is distributed throughout.Understanding the properties of a hollow sphere is crucial in physics because:- It determines the specific moment of inertia \( I_c = \frac{2}{3}MR^2 \).- The hollow nature affects how forces, like gravitational pull, influence rotation and swinging.This design means that the sphere's movement when suspended and used as a pendulum will differ from other shapes or configurations. The hollow sphere's peculiarities must be accounted for when solving physics problems involving moment of inertia and pendulum dynamics.

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Most popular questions from this chapter

In a physics lab, you attach a \(0.200-\mathrm{kg}\) air-track glider to the end of an ideal spring of negligible mass and start it oscillating. The elapsed time from when the glider first moves through the equilibrium point to the second time it moves through that point is 2.60 s. Find the spring's force constant.

An apple weighs 1.00 \(\mathrm{N}\) . When you hang it from the end of a long spring of force constant 1.50 \(\mathrm{N} / \mathrm{m}\) and negligible mass, it bounces up and down in SHM. If you stop the bouncing and let the apple swing from side to side through a small angle, the frequency of this simple pendulum is half the bounce frequency. (Because the angle is small, the back- and-forth swings do not cause any appreciable change in the length of the spring.) What is the unstretched length of the spring (with the apple removed)?

A \(1.50-\mathrm{kg},\) horizontal, uniform tray is attached to a vertical ideal spring of force constant 185 \(\mathrm{N} / \mathrm{m}\) and a \(275-\mathrm{g}\) metal ball is in the tray. The spring is below the tray, so it can oscillate up-and-down. The tray is then pushed down 15.0 \(\mathrm{cm}\) below its equilibrium point (call this point \(A\) ) and released from rest. (a) How high above point \(A\) will the tray be when the metal ball leaves the tray? (Hint: This does not oceur when the ball and tray reach their maximum speeds.) (b) How much time elapses between releasing the system at point \(A\) and the ball leaving the tray? (c) How fast is the ball moving just as it leaves the tray?

After landing on an unfamiliar planet, a space explorer con- structs a simple pendulum of length 50.0 \(\mathrm{cm} .\) She finds that the pendulum makes 100 complete swings in 136 \(\mathrm{s}\) . What is the value of \(g\) on this planet?

A rocket is accelerating upward at 4.00 \(\mathrm{m} / \mathrm{s}^{2}\) from the launchpad on the earth. Inside a small \(1.50-\mathrm{kg}\) ball hangs from the ceiling by a light \(1.10-\mathrm{m}\) wire. If the ball is displaced \(8.50^{\circ}\) from the vertical and released, find the amplitude and period of the resulting swings of this pendulum.

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