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A thin metal disk with mass \(2.00 \times 10^{-3} \mathrm{~kg}\) and radius \(2.20 \mathrm{~cm}\) is attached at its center to a long fiber (Fig. \(\mathbf{E 1 4 . 4 0}\) ). The disk, when twisted and released, oscillates with a period of \(1.00 \mathrm{~s}\). Find the torsion constant of the fiber.

Short Answer

Expert verified
The torsion constant of the fiber is \(1.925 \times 10^{-5}\) Nm/rad.

Step by step solution

01

Calculate the Moment of Inertia

First, let's use the formula for the moment of inertia of a circular disk. This formula is \(I = \frac{1}{2}mr^2\), where \(m = 2.00\times10^{-3}\) kg is the mass of the disk and \(r = 2.20\) cm = 0.022 m is the radius. Substituting the given values, the moment of inertia will be \(I = \frac{1}{2} \times 2.00\times10^{-3} \times (0.022)^2 = 4.84\times10^{-8}\) kg·m².
02

Calculate the Torsion Constant

The period of oscillation \(T\) is given by \(1.00\) s. Since the formula of the period of a torsion oscillator is \(T = 2\pi \sqrt{\frac{I}{\kappa}}\), we can rearrange it to express the torsion constant \(\kappa\): \(\kappa=\frac{I}{(T/(2\pi))^2}\). Now substitute the values: \(\kappa=\frac{4.84\times10^{-8}}{(1/(2\pi))^2} = 1.925 \times 10^{-5}\) Nm/rad.

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Key Concepts

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

Moment of Inertia
The moment of inertia is a fundamental concept in physics, often referred to as the "rotational inertia" of an object. Think of it as the resistance a body has to change its rotational motion. It's the rotational equivalent of mass in linear motion. When dealing with disk-like shapes, the formula for calculating moment of inertia is crucial: \(I = \frac{1}{2}mr^2\).
In this formula:
  • \(m\) represents the mass of the object.
  • \(r\) stands for the radius.
  • \(I\) results in the moment of inertia, expressed in kg·m².
This formula shows that the moment of inertia increases with more mass and a larger radius. It's especially relevant when analyzing how objects rotate around an axis. For our disk, with a given mass of \(2.00 \times 10^{-3}\) kg and radius \(2.20\) cm, the calculated moment of inertia is approximately \(4.84\times10^{-8}\) kg·m².
This simple but important formula gives us insight into the object's resistance to rotational changes.
Torsion Oscillator
A torsion oscillator is a system in which a disk or rod experiences rotational motion when twisted and released. It's quite like a pendulum, but for rotational instead of linear movement. When the disk is twisted, the fiber attached to it works like a spring, trying to return to its untwisted state, causing the disk to oscillate back and forth around its equilibrium position.
In such a system, the torsion constant (\(\kappa\)) plays a fundamental role. This constant is analogous to the spring constant in linear motion. It quantifies how stiff or resilient the fiber is when twisted. The larger the torsion constant, the stiffer the fiber, meaning it requires more torque to twist it.
By applying the formula for the period of a torsion oscillator, \(T = 2\pi \sqrt{\frac{I}{\kappa}}\), we can use known values of period and moment of inertia to find the torsion constant. For our disk-fiber setup, with a period of \(1.00\) second, the torsion constant turns out to be \(1.925 \times 10^{-5}\) Nm/rad. Understanding this value helps us comprehend the dynamics of the oscillating system.
Oscillation Period
The oscillation period represents the time it takes for the disk to complete one full back-and-forth twist. It's a crucial characteristic of any oscillatory motion, indicating not just the speed of oscillation but also reflecting the physical properties of the oscillator itself.
For torsion oscillators, the period \(T\) is linked to both the moment of inertia \(I\) and the torsion constant \(\kappa\). The mathematical expression \(T = 2\pi \sqrt{\frac{I}{\kappa}}\) is fundamental here.
  • A larger moment of inertia means a longer period, as there's more rotational resistance.
  • A stiffer fiber (larger \(\kappa\)) results in a shorter period, twisting more quickly back to its original position.
Understanding the period's relationship with these variables is key when studying rotational dynamics and designing systems where timing and stability of oscillations are important.

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Most popular questions from this chapter

A block of mass \(m\) is undergoing SHM on a horizontal, friction less surface while attached to a light, horizontal spring. The spring has force constant \(k,\) and the amplitude of the motion of the block is \(A\). (a) The average speed is the total distance traveled by the block divided by the time it takes it to travel this distance. Calculate the average speed for one cycle of the SHM. (b) How does the average speed for one cycle compare to the maximum speed \(v_{\max } ?\) (c) Is the average speed more or less than half the maximum speed? Based on your answer, does the block spend more time while traveling at speeds greater than \(v_{\max } / 2\) or less than \(v_{\max } / 2 ?\)

A Pendulum on Mars. A certain simple pendulum has a period on the earth of 1.60 s. What is its period on the surface of Mars, where \(g=3.71 \mathrm{~m} / \mathrm{s}^{2} ?\)

A rifle bullet with mass \(8.00 \mathrm{~g}\) and initial horizontal velocity \(280 \mathrm{~m} / \mathrm{s}\) strikes and embeds itself in a block with mass \(0.992 \mathrm{~kg}\) that rests on a friction less surface and is attached to one end of an ideal spring. The other end of the spring is attached to the wall. The impact compresses the spring a maximum distance of \(15.0 \mathrm{~cm} .\) After the impact, the block moves in SHM. Calculate the period of this motion.

A \(5.00 \mathrm{~kg}\) partridge is suspended from a pear tree by an ideal spring of negligible mass. When the partridge is pulled down \(0.100 \mathrm{~m}\) below its equilibrium position and released, it vibrates with a period of \(4.20 \mathrm{~s}\). (a) What is its speed as it passes through the equilibrium position? (b) What is its acceleration when it is \(0.050 \mathrm{~m}\) above the equilibrium position? (c) When it is moving upward, how much time is required for it to move from a point \(0.050 \mathrm{~m}\) below its equilibrium position to a point \(0.050 \mathrm{~m}\) above it? (d) The motion of the partridge is stopped, and then it is removed from the spring. How much does the spring shorten?

\(\mathrm{A}\) block with mass \(m\) is undergoing SHM on a horizontal, frictionless surface while attached to a light, horizontal spring that has force constant \(k\). You use motion sensor equipment to measure the maximum speed of the block during its oscillations. You repeat the measurement for the same spring and blocks of different masses while keeping the amplitude \(A\) at a constant value of \(12.0 \mathrm{~cm}\). You plot your data as \(v_{\max }^{2}\) versus \(1 / m\) and find that the data lie close to a straight line that has slope \(8.62 \mathrm{~N} \cdot \mathrm{m} .\) What is the force constant \(k\) of the spring?

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