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Chapter 9: Problem 4

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

Short Answer

Expert verified
(a) \( \frac{10\pi}{9} \) rad/s, (b) 1.8 seconds.

Step by step solution

01

Convert Revolutions to Radians

To find the angular speed in radians per second, we first need to convert revolutions per minute (rpm) to radians per minute. Since one revolution is equal to \( 2\pi \) radians, we multiply 33 1/3 rpm by \( 2\pi \) to get the angular distance in radians per minute: \( 33\frac{1}{3} \times 2\pi = \frac{100}{3} \times 2\pi = \frac{200\pi}{3} \) radians per minute.
02

Convert Minutes to Seconds

Now, we need to convert the angular speed from radians per minute to radians per second. There are 60 seconds in one minute, so we divide the angular speed by 60: \( \frac{200\pi}{3} \div 60 = \frac{200\pi}{180} = \frac{10\pi}{9} \) radians per second. This is the angular speed we are looking for in part (a).
03

Find the Period

The period is the time it takes to complete one full revolution. We start by finding the number of revolutions per second by dividing the revolutions per minute by 60: \( 33\frac{1}{3} \div 60 = \frac{1}{1.8} = \frac{3}{5.4} = \frac{3}{3.2} \approx 0.5556 \) revolutions per second. The period \( T \) is the reciprocal of this frequency, so \( T \approx \frac{1}{0.5556} \approx 1.8 \) seconds.

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Key Concepts

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

LP Records
LP records, known as long-play records, revolutionized the music industry in the mid-20th century. These records are 12 inches in diameter and were a popular medium for music listening. One unique feature was their rotational speed, spinning at 33 1/3 revolutions per minute (RPM). LP records could hold more music due to the way they were manufactured. This allowed artists to feature entire albums on a single disc. It became a cultural icon, influencing music production and distribution before the digital era began. Understanding the mechanics of LPs involves engaging with concepts like angular speed, as we often calculate how fast the records spin when played.
Radians Conversion
To discuss angular speed, we need to understand radians. A radian is a unit of angular measure used in mathematics. In terms of circular motion, one complete revolution is equal to the angle produced by the radius creating the circumference of a circle.A full circle in radians is equal to \(2\pi\) radians. This conversion is crucial when dealing with revolving bodies, like LP records. We often need to transform revolutions into radians, as radians offer a practical unit for describing angular velocity.Radians make calculations involving angles and rotation straightforward, since \(2\pi\) simplifies the relationship between a circle鈥檚 radius and its circumference.
Revolutions Per Minute (RPM)
Revolutions per minute (RPM) is a common unit of rotational speed. It tells us how many complete turns an object makes in one minute. For LP records, a standard speed of 33 1/3 RPM meant the record turned precisely 33.33 times in one minute. To calculate angular speed in radians per second, understanding RPM is essential. It's the starting point for conversions and calculations needed to find an LP record鈥檚 angular speed in different units. Using RPM, one can determine aspects like how fast a record spins or how long it takes to complete a revolution, making it helpful for solving problems in physics and engineering related to rotational dynamics.
Revolutions to Radians Conversion
Converting revolutions to radians is a key step in determining angular speed. Since one revolution is equivalent to \(2\pi\) radians, to find the angular speed in radians per minute for an LP spinning at 33 1/3 RPM, we multiply \(33\frac{1}{3}\) by \(2\pi\).Subsequently, converting this value from radians per minute to radians per second involves dividing by 60, because there are 60 seconds in a minute. In our exercise, this results in an angular speed of \(\frac{10\pi}{9}\) radians per second.This conversion is not just essential for understanding LP records but also for analyzing any object that rotates, making radians a fundamental aspect of rotational motion studies.

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Most popular questions from this chapter

A \(392 \mathrm{~N}\) wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at \(25.0 \mathrm{rad} / \mathrm{s} .\) The radius of the wheel is \(0.600 \mathrm{~m},\) and its moment of inertia about its rotation axis is \(0.800 \mathrm{MR}^{2}\). Friction does work on the wheel as it rolls up the hill to a stop, a height \(h\) above the bottom of the hill; this work has absolute value 3500 J. Calculate \(h\).

A \(7300 \mathrm{~N}\) elevator is to be given an acceleration of \(0.150 \mathrm{~g}\) by connecting it to a cable of negligible weight wrapped around a turning cylindrical shaft. If the shaft's diameter can be no larger than 16.0 \(\mathrm{cm}\) due to space limitations, what must be its minimum angular acceleration to provide the required acceleration of the elevator?

The spin cycles of a washing machine have two angular speeds, 423 rev \(/ \mathrm{min}\) and \(640 \mathrm{rev} / \mathrm{min} .\) The internal diameter of the drum is \(0.470 \mathrm{~m}\). (a) What is the ratio of the maximum radial force on the laundry for the higher angular speed to that for the lower speed? (b) What is the ratio of the maximum tangential speed of the laundry for the higher angular speed to that for the lower speed? (c) Find the laundry's maximum tangential speed and the maximum radial acceleration, in terms of \(g\).

It has been suggested that we should use our power plants to generate energy in the off-hours (such as late at night) and store it for use during the day. One idea put forward is to store the energy in large flywheels. Suppose we want to build such a flywheel in the shape of a hollow cylinder of inner radius \(0.500 \mathrm{~m}\) and outer radius \(1.50 \mathrm{~m},\) using concrete of density \(2.20 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\). (a) If, for stability, such a heavy flywheel is limited to 1.75 seconds for each revolution and has negligible friction at its axle, what must be its length to store \(2.5 \mathrm{MJ}\) of energy in its rotational motion? (b) Suppose that by strengthening the frame you could safely double the flywheel's rate of spin. What length of flywheel would you need in that case? (Solve this part without reworking the entire problem!)

BIO The Spinning Eel. American eels are freshwater fish with long, slender bodies that we can treat as a uniform cylinder \(1.0 \mathrm{~m}\) long and 10 \(\mathrm{cm}\) in diameter. An eel compensates for its small jaw and teeth by holding onto prey with its mouth and then rapidly spinning its body around its long axis to tear off a piece of flesh. Eels have been recorded to spin at up to \(14 \mathrm{rev} / \mathrm{s}\) when feeding in this way. Although this feeding method is energetically costly, it allows the eel to feed on larger prey than it otherwise could. A field researcher uses the slow-motion feature on her phone's camera to shoot a video of an eel spinning at its maximum rate. The camera records at 120 frames per second. Through what angle does the eel rotate from one frame to the next? A. \(1^{\circ}\) B. \(10^{\circ}\) C. \(22^{\circ}\) D. \(42^{\circ}\)
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