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31\. Balance Wheel The balance wheel of a watch oscillates with a rotational amplitude of \(\pi\) rad and a period of \(0.500 \mathrm{~s}\) Find (a) the maximum rotational speed of the wheel, (b) the rotational speed of the wheel when its displacement is \(\pi / 2 \mathrm{rad}\), and (c) the magnitude of the rotational acceleration of the wheel when its displacement is \(\pi / 4 \mathrm{rad}\).

Short Answer

Expert verified
(a) \text{4\pi^2 rad/s}, (b) 2\pi^2\sqrt{3} rad/s, (c) \-4\pi^3 rad/s^2

Step by step solution

01

Determine the angular frequency

The angular frequency \(\textbf{\omega}\) can be found using the period \(\textbf{T} = 0.500 \mathrm{s}\) with the formula \[ \omega = \frac{2\pi}{T} \]. Substitute \(\textbf{T}\) into the equation. \[ \omega = \frac{2\pi}{0.500} = 4\pi \text{ rad/s} \]
02

Calculate the maximum rotational speed

The maximum rotational speed \( \textbf{\omega_{max}} \) is given by \[ \omega_{max} = \omega A \] where \(A\) is the amplitude. Given \(A = \pi\) rad, we have \[ \omega_{max} = (4\pi)(\pi) = 4\pi^2 \text{ rad/s} \]
03

Find the rotational speed at \( \pi/2 \text{ rad} \)

Use the energy conservation relation for simple harmonic motion: \(\textbf{\omega_x = \omega \sqrt{A^2 - x^2}}\). Substituting \(\textbf{x = \pi/2}\) and \(\textbf{A = \pi}\), \[\omega_x = 4\pi \sqrt{\pi^2 - (\pi/2)^2} = 4\pi \sqrt{\pi^2 - \pi^2/4} = 4\pi \sqrt{3\pi^2/4} = 4\pi (\frac{\pi\sqrt{3}}{2}) = 2\pi^2\sqrt{3} \text{ rad/s} \]
04

Find the rotational acceleration at \(\ \pi/4 \text{ rad}\)

The rotational acceleration \(\textbf{\alpha}\) is given by \[-\omega^2 x\], where \(\omega = 4\pi\) and \(x = \pi/4\). Therefore, \(\textbf{\alpha}\) is given by: \[ \alpha = -(4\pi)^2 \left(\frac{\pi}{4}\right) = -16\pi^2\frac{\pi}{4} = -4\pi^3 \text{ rad/s}^2 \]

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Key Concepts

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

Angular Frequency
In simple harmonic motion, one important concept is the angular frequency, denoted by \(\textbf{\omega}\). This tells us how many radians the object moves through per second. Angular frequency is related to the period \(\textbf{T}\) of the oscillation by the formula: \[ \omega = \frac{2\pi}{T} \] The period is the time it takes for one complete cycle of motion. So, using the given period \(0.500 \text{ s}\), we substitute it into the formula to find \(\textbf{\omega}\): \[ \omega = \frac{2\pi}{0.500} = 4\pi \text{ rad/s} \] This means that every second, the watch's balance wheel moves through \(4\pi\) radians.
Rotational Speed
The rotational speed of an oscillating object shows how quickly it moves at any given point in time. The maximum rotational speed, \( \textbf{\omega_{max}} \), occurs at the equilibrium position (where the displacement is zero). It can be calculated using the formula: \[ \omega_{max} = \omega A \] Here, \(A\) is the amplitude of motion. Given \(A = \pi\text{ rad}\) and \( \omega = 4\pi \text{ rad/s}\), we have: \[ \omega_{max} = (4\pi)(\pi) = 4\pi^2 \text{ rad/s} \] This is the fastest the wheel moves. To find the rotational speed when the displacement is \(\pi / 2 \text{ rad}\), we use the energy conservation relation: \[ \omega_x = \omega \sqrt{A^2 - x^2} \] With \( x = \pi / 2 \text{ rad} \) and \( A = \pi \text{ rad} \), we get: \[ \omega_x = 4\pi \sqrt{\pi^2 - (\pi / 2)^2} = 4\pi \sqrt{\pi^2 - \pi^2 / 4} = 4\pi \sqrt{3\pi^2 / 4} = 4\pi (\frac{\pi\sqrt{3}}{2}) = 2\pi^2\sqrt{3} \text{ rad/s} \] This formula provides the rotational speed at different points of the oscillation.
Rotational Acceleration
Rotational acceleration, represented by \(\textbf{\alpha}\), shows how quickly the rotational speed changes. It can be determined using the formula: \[ \alpha = -\omega^2 x \] Negative sign indicates that the acceleration is directed opposite to the displacement. Using \( \omega = 4\pi \text{ rad/s} \) and displacement \( x = \pi / 4 \text{ rad} \): \[ \alpha = -(4\pi)^2 (\pi / 4) = -16\pi^2(\pi / 4) = -4\pi^3 \text{ rad/s}^2 \] This formula gives the magnitude of rotational acceleration at a specific point in simple harmonic motion.

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Most popular questions from this chapter

32\. Flat Disk A flat uniform circular disk has a mass of \(3.00 \mathrm{~kg}\) and a radius of \(70.0 \mathrm{~cm}\). It is suspended in a horizontal plane by a vertical wire attached to its center. If the disk is rotated \(2.50\) rad about the wire, a torque of \(0.0600 \mathrm{~N} \cdot \mathrm{m}\) is required to maintain that orientation. Calculate (a) the rotational inertia of the disk about the wire, (b) the torsion constant, and (c) the angular frequency of this torsion pendulum when it is set oscillating.

22\. Two Particles Two particles execute simple harmonic motion of the same amplitude and frequency along close parallel lines. They pass each other moving in opposite directions each time their displacement is half their amplitude. What is their phase difference?

56\. Vertical Spring A vertical spring stretches \(9.6 \mathrm{~cm}\) when a \(1.3 \mathrm{~kg}\) block is hung from its end. (a) Calculate the spring constant. This block is then displaced an additional \(5.0 \mathrm{~cm}\) downward and released from rest. Find (b) the period, (c) the frequency, (d) the amplitude, and (e) the maximum speed of the resulting SHM.

51\. Horizontal Frictionless A \(5.00 \mathrm{~kg}\) object on a horizontal frictionless surface is attached to a spring with spring constant 1000 \(\mathrm{N} / \mathrm{m} .\) The object is displaced from equilibrium \(50.0 \mathrm{~cm}\) horizontally and given an initial velocity of \(10.0 \mathrm{~m} / \mathrm{s}\) back toward the equilibrium position. (a) What is the frequency of the motion? What are (b) the initial potential energy of the block-spring system, (c) the initial kinetic energy, and (d) the hmolitude of the oscillation?

62\. Washboard Road A \(1000 \mathrm{~kg}\) car carrying four \(82 \mathrm{~kg}\) people travels over a rough "washboard" dirt road with corrugations \(4.0 \mathrm{~m}\) apart, which cause the car to bounce on its spring suspension. The car bounces with maximum amplitude when its speed is \(16 \mathrm{~km} / \mathrm{h}\). The car now stops, and the four people get out. By how much does the car body rise on its suspension due to this decrease in mass?

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