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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 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
To calculate the period of the hollow sphere physical pendulum, use the formula for the period of a physical pendulum, the characteristics of the sphere, and the Parallel Axis Theorem to find the Moment of Inertia.

Step by step solution

01

Formula of a Physical Pendulum Period

The formula for the period of a physical pendulum is given by \(T = 2\pi \sqrt{\frac{I}{mgh}}\), where \(T\) is the period, \(m\) is the mass of the pendulum, \(g\) is the acceleration due to gravity, \(h\) is the height of the center of mass above the pivot, and \(I\) is the Moment of Inertia.
02

Moment of Inertia via Parallel Axes Theorem

The Parallel Axis Theorem can be used to calculate the moment of inertia of the ornament. It states that \(I = I_{\text_{cm}} + MD^2\), where \(I_{\text_{cm}}\) is the moment of inertia of the body about an axis passing through the center of mass and parallel to the given axis, \(M\) is the total mass of the body, and \(D\) is the distance between the two parallel axes, which is equal to the radius \(R\) in this case. For a hollow sphere, \(I_{\text_{cm}}\) is \(2/3 MR^2\). Therefore, \(I = 2/3 Mr^2 + Mr^2 = 5/3 Mr^2\).
03

Find Height of Center of Mass

The height of the center of mass above the pivot is simply the radius \(R\) of the sphere.
04

Substitute Values and Solve

Finally, substitute the values into the formula in Step 1 with \(m=0.015 kg\), \(R=0.05 m\), \(g=9.81 m/s^2\), and \(I=5/3 Mr^2\). Simplifying gives the period \(T\).

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Key Concepts

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

Parallel-Axis Theorem
Understanding the parallel-axis theorem is crucial when calculating the moment of inertia for an object rotating about an axis that is not through its center of mass. This mathematical tool allows us to relate the moment of inertia about an axis through the center of mass (\(I_{\text{cm}}\text)\) to the moment of inertia about any parallel axis.The theorem states that the moment of inertia about the parallel axis (\(I\text)\) can be found using the formula \[\begin{equation} I = I_{\text{cm}} + MD^2 \end{equation}\]where
  • \(I_{\text{cm}}\text)\) is the moment of inertia about the center of mass,
  • \(M\text)\) is the mass of the object,
  • \(D\text)\) is the perpendicular distance between the two axes.
When applied to real-world problems, like calculating the swing of a holiday ornament or any other physical pendulum, the parallel-axis theorem is indispensable for determining how the distribution of mass around the pivot point affects the pendulum's motion.
Moment of Inertia
The moment of inertia is a measure of an object's resistance to changes in its rotational motion. In simple terms, it's akin to mass in linear motion, but for rotation. The value of the moment of inertia depends on how the object's mass is distributed in relation to the axis of rotation.In the case of our holiday ornament, which is a hollow sphere, the moment of inertia about an axis through the center of mass is given by the formula\[\begin{equation} I_{\text{cm}} = \frac{2}{3}MR^2 \end{equation}\]Here, \(M\text)\) is the mass of the sphere, and \(R\text)\) is its radius. When the axis of rotation is moved away from the center of mass, such as when the sphere swings as a pendulum, the parallel-axis theorem must be used to calculate the new moment of inertia, involving both the sphere's geometry and the distance from the new axis to the center of mass.
Simple Harmonic Motion
Simple harmonic motion (SHM) describes a type of predictable, oscillating movement found in systems where the restoring force is directly proportional to the displacement and acts in the opposite direction. This is typical for pendulums and springs.

Characteristics of SHM

  • It is periodic, meaning it repeats in cycles.
  • It has a fixed frequency that depends on the system's properties, such as mass and stiffness for a spring or length and gravity for a pendulum.
  • The motion is sinusoidal, following sine or cosine functions over time.
The period of a physical pendulum exhibiting SHM, like the holiday ornament, is calculated using a formula related to the system's moment of inertia and the gravitational torque. For our ornament, it represents the time taken for one complete back and forth swing.
Center of Mass
The center of mass of an object is the point where its mass is considered to be concentrated for the purpose of analysis. In the context of rotational motion, it plays a pivotal role as it's the point about which the object's mass is evenly distributed.When dealing with the motion of rigid bodies, like our physical pendulum (the holiday ornament), its center of mass hangs directly below the pivot point when at rest. In calculating the period of oscillation, the height above the pivot point to the center of mass (\(h\text)\) is a crucial variable. Since the ornament hangs from a wire, the center of mass does not shift from the center of the sphere, thereby simplifying the calculation process. The entire mass of the ornament can be thought to act through this single point during its swinging motion.

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Most popular questions from this chapter

BIO Weighing Astronauts. This procedure has been used to "weigh" astronauts in space: A \(42.5 \mathrm{~kg}\) chair is attached to a spring and allowed to oscillate. When it is empty, the chair takes \(1.30 \mathrm{~s}\) to make one complete vibration. But with an astronaut sitting in it, with her feet off the floor, the chair takes \(2.54 \mathrm{~s}\) for one cycle. What is the mass of the astronaut?

A small sphere with mass \(m\) is attached to a massless rod of length \(L\) that is pivoted at the top, forming a simple pendulum. The pendulum is pulled to one side so that the rod is at an angle \(\theta\) from the vertical, and released from rest. (a) In a diagram, show the pendulum just after it is released. Draw vectors representing the forces acting on the small sphere and the acceleration of the sphere. Accuracy counts! At this point, what is the linear acceleration of the sphere? (b) Repeat part (a) for the instant when the pendulum rod is at an angle \(\theta / 2\) from the vertical. (c) Repeat part (a) for the instant when the pendulum rod is vertical. At this point, what is the linear speed of the sphere?

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. When the amplitude of the motion is \(0.090 \mathrm{~m},\) it takes the block \(2.70 \mathrm{~s}\) to travel from \(x=0.090 \mathrm{~m}\) to \(x=-0.090 \mathrm{~m} .\) If the amplitude is doubled, to \(0.180 \mathrm{~m},\) how long does it take the block to travel (a) from \(x=0.180 \mathrm{~m}\) to \(x=-0.180 \mathrm{~m}\) and (b) from \(x=0.090 \mathrm{~m}\) to \(x=-0.090 \mathrm{~m} ?\)

The jerk is defined to be the time rate of change of the acceleration. (a) If the velocity of an object undergoing SHM is given by \(v_{x}=-\omega A \sin (\omega t),\) what is the equation for the \(x\) -component of the jerk as a function of time? (b) What is the value of \(x\) for the object when the \(x\) -component of the jerk has its largest positive value? (c) What is \(x\) when the \(x\) -component of the jerk is most negative? (d) When it is zero? (e) If \(v_{x}\) equals \(-0.040 \mathrm{~s}^{2}\) times the \(x\) -component of the jerk for all \(t,\) what is the period of the motion?

Two uniform solid spheres, each with mass \(M=0.800 \mathrm{~kg}\) and radius \(R=0.0800 \mathrm{~m},\) are connected by a short, light rod that is along a diameter of each sphere and are at rest on a horizontal tabletop. A spring with force constant \(k=160 \mathrm{~N} / \mathrm{m}\) has one end attached to the wall and the other end attached to a friction less ring that passes over the rod at the center of mass of the spheres, which is midway between the centers of the two spheres. The spheres are each pulled the same distance from the wall, stretching the spring, and released. There is sufficient friction between the tabletop and the spheres for the spheres to roll without slipping as they move back and forth on the end of the spring. Show that the motion of the center of mass of the spheres is simple harmonic and calculate the period.

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