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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/sĀ²), 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 secondsā€”a 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

An object is undergoing SHM with period 0.300 s and amplitude 6.00 cm. At \(t =\) 0 the object is instantaneously at rest at \(x =\) 6.00 cm. Calculate the time it takes the object to go from \(x =\) 6.00 cm to \(x = -\)1.50 cm.

While on a visit to Minnesota ("Land of 10,000 Lakes"), you sign up to take an excursion around one of the larger lakes. When you go to the dock where the 1500-kg boat is tied, you find that the boat is bobbing up and down in the waves, executing simple harmonic motion with amplitude 20 cm. The boat takes 3.5 s to make one complete up-and-down cycle. When the boat is at its highest point, its deck is at the same height as the stationary dock. As you watch the boat bob up and down, you (mass 60 kg) begin to feel a bit woozy, due in part to the previous night's dinner of lutefisk. As a result, you refuse to board the boat unless the level of the boat's deck is within 10 cm of the dock level. How much time do you have to board the boat comfortably during each cycle of up-and-down motion?

At the end of a ride at a winter-theme amusement park, a sleigh with mass 250 kg (including two passengers) slides without friction along a horizontal, snow-covered surface. The sleigh hits one end of a light horizontal spring that obeys Hooke's law and has its other end attached to a wall. The sleigh latches onto the end of the spring and subsequently moves back and forth in SHM on the end of the spring until a braking mechanism is engaged, which brings the sleigh to rest. The frequency of the SHM is 0.225 Hz, and the amplitude is 0.950 m. (a) What was the speed of the sleigh just before it hit the end of the spring? (b) What is the maximum magnitude of the sleigh's acceleration during its SHM?

A 0.500-kg glider, attached to the end of an ideal spring with force constant \(k =\) 450 N/m, undergoes SHM with an amplitude of 0.040 m. Compute (a) the maximum speed of the glider; (b) the speed of the glider when it is at \(x = -\)0.015 m; (c) the magnitude of the maximum acceleration of the glider; (d) the acceleration of the glider at \(x = -\)0.015 m; (e) the total mechanical energy of the glider at any point in its motion.

The point of the needle of a sewing machine moves in SHM along the \(x\)-axis with a frequency of 2.5 Hz. At \(t =\) 0 its position and velocity components are \(+\)1.1 cm and \(-\)15 cm/s, respectively. (a) Find the acceleration component of the needle at \(t =\) 0. (b) Write equations giving the position, velocity, and acceleration components of the point as a function of time.

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