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Chapter 9: Problem 4

\(\cdot\) The once-popular LP (long-play) records were 12 in. in diameter and turned at a constant 33\(\frac{1}{3}\) rpm. Find (a) the angular speed of the LP in rad/s and (b) its period in seconds.

Short Answer

Expert verified
(a) Angular speed is \(\frac{10\pi}{9}\) rad/s; (b) Period is 1.8 seconds.

Step by step solution

01

Understand the Problem

The problem asks us to find two things: (a) the angular speed of the LP in radians per second and (b) the period of rotation in seconds. The LP turns at 33\(\frac{1}{3}\) revolutions per minute.
02

Convert Revolutions per Minute to Revolutions per Second

First, convert the angular speed from revolutions per minute (rpm) to revolutions per second (rps).To do this, divide 33\(\frac{1}{3}\) by 60.\[ 33\frac{1}{3} = 33 + \frac{1}{3} = \frac{100}{3} \]Now divide by 60 to convert to revolutions per second:\[ \frac{100}{3} \div 60 = \frac{100}{3 \times 60} = \frac{100}{180} = \frac{5}{9} \text{ rps} \]
03

Convert Revolutions per Second to Radians per Second

Since one revolution is equivalent to \(2\pi\) radians, multiply the revolutions per second by \(2\pi\) to convert to radians per second.\[\text{Angular speed} = \frac{5}{9} \times 2\pi = \frac{10\pi}{9} \text{ rad/s} \]
04

Calculate the Period of Rotation

The period \(T\) is the time taken for one complete rotation. It is the reciprocal of the angular speed in revolutions per second.Since we found the angular speed in revolutions per second as \(\frac{5}{9}\) rps, the period is:\[T = \frac{1}{\frac{5}{9}} = \frac{9}{5} \approx 1.8 \text{ seconds} \]
05

Review the Results

Therefore, the angular speed of the LP in radians per second is \(\frac{10\pi}{9}\), and the period of rotation is approximately 1.8 seconds.

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Key Concepts

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

Radian
When exploring angular displacement, the concept of a radian becomes essential. A radian is a measure of angle based on the radius of a circle. It offers a different perspective from the common degree measurement and simplifies the mathematics involved in circular motion.
A radian is defined as the angle created when the arc length is equal to the radius of the circle.
Since the circumference of a circle is given by the formula \(2\pi r\), where \(r\) is the radius, there are \(2\pi\) radians in a complete circle.
This means:
  • 180掳 is equivalent to \(\pi\) radians.
  • One complete revolution (360掳) corresponds to \(2\pi\) radians.
Using radians allows for more straightforward calculations in physics and engineering, particularly for continuous rotations such as those seen in LP records. When LPs spin on turntables, understanding radians helps translate those spins into a usable angular speed, as shown in exercises calculating revolutions to radians per second.
Revolution
In the context of rotational motion, a revolution represents one complete 360-degree cycle around a circle. It's a crucial concept when measuring how objects move in circular paths.
Revolutions help in expressing angular speed, which tells us how fast something spins.
For instance, an LP spinning at 33\(\frac{1}{3}\) revolutions per minute (rpm) indicates how many full turns it makes in one minute.
  • This can be converted to revolutions per second (rps) for applications requiring a second-based measurement.
  • The conversion is necessary for further calculation of angular speed in radians per second.
Knowing how many revolutions occur also aids in determining the period鈥攖he time for one complete cycle. As rotations relate directly to time, understanding this concept ties back to period calculations.
Period of Rotation
The period of rotation refers to the time needed for one complete revolution of an object moving in a circular path. It is often symbolized by \(T\) and is crucial in describing motion in applications like spinning LPs.
The period is inversely related to the angular speed, meaning if we know how quickly something spins, we can find out how long it takes to complete one spin.
  • If the LP turns at \(\frac{5}{9}\) revolutions per second, this means that each full cycle or rotation takes \( \frac{9}{5} \approx 1.8 \) seconds.
  • Calculating the period involves taking the reciprocal of the spins per second.
Understanding the period is essential for timing and sequencing in processes dependent on cyclical motions, like music playback on a turntable.

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Most popular questions from this chapter

A solid uniform spherical stone starts moving from rest at the top of a hill. At the bottom of the hill the ground curves upward, launching the stone vertically a distance \(H\) below its start. How high will the stone go (a) if there is no friction on the hill and (b) if there is enough friction on the hill for the stone to roll without slipping? (c) Why do you get two different answers even though the stone starts with the same gravitational potential energy in both cases?

A solid uniform marble and a block of ice, each with the same mass, start from rest at the same height \(H\) above the bottom of a hill and move down it. The marble rolls without slipping, but the ice slides without friction. (a) Find the speed of each of these objects when it reaches the bottom of the hill. (b) Which object is moving faster at the bottom, the ice or the marble? (c) Which object has more kinetic energy at the bottom, the ice or the marble?

A A flywheel having constant angular acceleration requires 4.00 s to rotate through 162 rad. Its angular velocity at the end of this time is 108 rad/s. Find (a) the angular velocity at the beginning of the 4.00 s interval; (b) the angular acceleration of the flywheel.

\(\bullet\) A size-5 soccer ball of diameter 22.6 \(\mathrm{cm}\) and mass 426 \(\mathrm{g}\) rolls up a hill without slipping, reaching a maximum height of 5.00 \(\mathrm{m}\) above the base of the hill. We can model this ball as a thin-walled hollow sphere. (a) At what rate was it rotating at the base of the hill? (b) How much rotational kinetic energy did it then have?

\(\bullet\) A flywheel with a radius of 0.300 \(\mathrm{m}\) starts from rest and accelerates with a constant angular acceleration of 0.600 \(\mathrm{rad} / \mathrm{s}^{2} .\) Compute the magnitude of the tangential acceleration, the radial acceleration, and the resultant acceleration of a point on its rim (a) at the start, (b) after it has turned through \(60.0^{\circ},\) and \((\mathrm{c})\) after it has turned through \(120.0^{\circ} .\)
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