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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

\(\mathrm{A} 50.0 \mathrm{~g}\) hard-boiled egg moves on the end of a spring with force constant \(k=25.0 \mathrm{~N} / \mathrm{m} .\) Its initial displacement is \(0.300 \mathrm{~m} . \mathrm{A}\) damping force \(F_{x}=-b v_{x}\) acts on the egg, and the amplitude of the motion decreases to \(0.100 \mathrm{~m}\) in \(5.00 \mathrm{~s}\). Calculate the magnitude of the damping constant \(b\).

A Spring with Mass. The preceding problems in this chapter have assumed that the springs had negligible mass. But of course no spring is completely massless. To find the effect of the spring's mass, consider a spring with mass \(M,\) equilibrium length \(L_{0},\) and spring constant \(k\). When stretched or compressed to a length \(L,\) the potential energy is \(\frac{1}{2} k x^{2},\) where \(x=L-L_{0}\). (a) Consider a spring, as described above, that has one end fixed and the other end moving with speed \(v\). Assume that the speed of points along the length of the spring varies linearly with distance \(l\) from the fixed end. Assume also that the mass \(M\) of the spring is distributed uniformly along the length of the spring. Calculate the kinetic energy of the spring in terms of \(M\) and \(v .\) (Hint: Divide the spring into pieces of length \(d l ;\) find the speed of each piece in terms of \(l, v,\) and \(L ;\) find the mass of each piece in terms of \(d l, M,\) and \(L ;\) and integrate from 0 to \(L .\) The result is \(n o t \frac{1}{2} M v^{2},\) since not all of the spring moves with the same speed.) (b) Take the time derivative of the conservation of energy equation, Eq. (14.21), for a mass \(m\) moving on the end of a massless spring. By comparing your results to Eq. (14.8), which defines \(\omega\), show that the angular frequency of oscillation is \(\omega=\sqrt{k / m}\). (c) Apply the procedure of part (b) to obtain the angular frequency of oscillation \(\omega\) of the spring considered in part (a). If the effective mass \(M^{\prime}\) of the spring is defined by \(\omega=\sqrt{k / M^{\prime}},\) what is \(M^{\prime}\) in terms of \(M ?\)

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 \(g=3.71 \mathrm{~m} / \mathrm{s}^{2} ?\)

Object \(A\) has mass \(m_{A}\) and is in \(\mathrm{SHM}\) on the end of a spring with force constant \(k_{A} .\) Object \(B\) has mass \(m_{B}\) and is in \(\mathrm{SHM}\) on the end of a spring with force constant \(k_{B}\). The amplitude \(A_{A}\) for object \(A\) is twice the amplitude \(A_{B}\) for the motion of object \(B\). Also, \(m_{B}=4 m_{A}\) and \(k_{A}=9 k_{B}\). (a) What is the ratio of the maximum speeds of the two objects, \(v_{\max , A} / v_{\max , B} ?\) (b) What is the ratio of their maximum accelerations, \(a_{\max , A} / a_{\max , B} ?\)

When an object of unknown mass is attached to an ideal spring with force constant \(120 \mathrm{~N} / \mathrm{m},\) it is found to vibrate with a frequency of \(6.00 \mathrm{~Hz}\). Find (a) the period of the motion; (b) the angular frequency; (c) the mass of the object.

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