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Chapter 9: Problem 44

You need to design an industrial turntable that is \(60.0 \mathrm{~cm}\) in diameter and has a kinetic energy of \(0.250 \mathrm{~J}\) when turning at \(45.0 \mathrm{rpm}\) (rev/min). (a) What must be the moment of inertia of the turntable about the rotation axis? (b) If your workshop makes this turntable in the shape of a uniform solid disk, what must be its mass?

Short Answer

Expert verified
By substituting the given information and solving the first two equations step-by-step, we obtain the necessary answers. (a) The moment of inertia of the turntable about the rotation axis is approximately \( 0.032 \, kg.m^2 \) and (b) the mass of this turntable is approximately \( 0.71 \, kg \). Round the answers as necessary for your specific requirement.

Step by step solution

01

Calculate the Moment of Inertia

From the formula of kinetic energy for rotating objects, which is \( \frac{1}{2} I 蠅^2 \), where \( I \) is the moment of inertia and \( 蠅 \) is the angular velocity, we rearrange for \( I \) thus: \( I = \frac{2K}{蠅^2} \). Given that the kinetic energy \( K = 0.250 J \) and the angular velocity \( 蠅 = 45.0 rpm = 45 \cdot \frac{2蟺}{60} rad/sec \), we substitute these values into the formula and solve for \( I \).
02

Calculate the mass of the turntable

For a uniformly solid disk of radius \( r \) and mass \( m \), the moment of inertia \( I \) can be calculated as \( \frac{1}{2} m r^2 \). Solving for mass \( m \) gives \( m = \frac{2I}{r^2} \). Substituting the calculated moment of inertia from the first step and the given radius \( r = \frac{60}{2} cm = 30 cm = 0.3 m \) into the formula will yield the mass of the turntable.

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Key Concepts

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

Understanding Kinetic Energy in Rotating Objects
Kinetic energy is typically associated with objects that are moving in a straight line, but objects in rotation possess kinetic energy too鈥攋ust in a slightly different form. For an object that rotates, such as the industrial turntable from our exercise, the kinetic energy is given by the formula

\[ K = \frac{1}{2} I \omega^2 \]
where:\[ K \] is the kinetic energy in joules (J), \[ I \] is the moment of inertia, and \[ \omega \] (omega) is the angular velocity in radians per second (rad/s). The moment of inertia represents how the mass is distributed in relation to the axis of rotation鈥攖he greater the distribution of mass from the center, the higher the moment of inertia, making it harder to spin the object. The kinetic energy in rotating objects like our turntable is a crucial factor in design because it affects how much energy is required to start or stop the rotation. Through this exercise, we see the direct relationship between an object's mass distribution, how fast it spins, and the energy it retains while rotating.
The role of Angular Velocity in Rotational Motion
Angular velocity is a measure of how quickly an object rotates about a specific point or axis, akin to how speed measures how fast something travels in a straight line. For our turntable, which rotates in revolutions per minute (rpm), we convert this to the standard unit of radians per second (rad/s) using the conversion factor \( 2\pi \) radians for a full revolution.

The conversion formula is:
\[ \omega = \text{angular speed in rpm} \times \frac{2\pi}{60} \]
Thus, a turntable spinning at 45.0 rpm translates to \( 45 \times \frac{2\pi}{60} \) rad/s. Angular velocity not only describes the speed of rotation but also plays a significant role in calculations of kinetic energy and moment of inertia for rotating objects. In our exercise, the angular velocity is squared in the energy formula, indicating that a small increase in rotational speed will result in a significant increase in kinetic energy.
Calculating the Mass of a Uniform Solid Disk
In the case of a solid disk, as with the industrial turntable we鈥檙e working on, the mass is uniformly distributed, and certain assumptions can be made for its calculation. The moment of inertia for a uniform solid disk about its center is given by

\[ I = \frac{1}{2} m r^2 \]
where \( m \) is the mass and \( r \) is the radius of the disk. When we need to find the mass of the disk, we rearrange the formula to solve for \( m \), giving us:
\[ m = \frac{2I}{r^2} \]
With the moment of inertia determined, as in the textbook solution, and the radius of the turntable provided, calculating the disk's mass becomes a matter of substituting these known values into the formula. This mass calculation is fundamental when designing objects like the turntable, as it influences the system's stability and the power requirements for the motor intended to drive it. A well-calculated mass ensures that the turntable performs efficiently under its intended operating conditions.

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Most popular questions from this chapter

At \(t=0\) the current to a dc electric motor is reversed, resulting in an angular displacement of the motor shaft given by \(\theta(t)=(250 \mathrm{rad} / \mathrm{s}) t-\left(20.0 \mathrm{rad} / \mathrm{s}^{2}\right) t^{2}-\left(1.50 \mathrm{rad} / \mathrm{s}^{3}\right) t^{3} .\) (a) At what time is the angular velocity of the motor shaft zero? (b) Calculate the angular acceleration at the instant that the motor shaft has zero angular velocity. (c) How many revolutions does the motor shaft turn through between the time when the current is reversed and the instant when the angular velocity is zero? (d) How fast was the motor shaft rotating at \(t=0,\) when the current was reversed? (e) Calculate the average angular velocity for the time period from \(t=0\) to the time calculated in part (a).

The Crab Nebula is a cloud of glowing gas about 10 light-years across, located about 6500 light-years from the earth (Fig. P9.86). It is the remnant of a star that underwent a supernova \(e x\) plosion, seen on earth in 1054 A.D. Energy is released by the Crab Nebula at a rate of about \(5 \times 10^{31} \mathrm{~W},\) about \(10^{5}\) times the rate at which the sun radiates energy. The Crab Nebula obtains its energy from the rotational kinetic energy of a rapidly spinning neutron star at its center. This object rotates once every \(0.0331 \mathrm{~s},\) and this period is increasing by \(4.22 \times 10^{-13} \mathrm{~s}\) for each second of time that elapses. (a) If the rate at which energy is lost by the neutron star is equal to the rate at which energy is released by the nebula, find the moment of inertia of the neutron star. (b) Theories of supernovae predict that the neutron star in the Crab Nebula has a mass about 1.4 times that of the sun. Modeling the neutron star as a solid uniform sphere, calculate its radius in kilometers. (c) What is the linear speed of a point on the equator of the neutron star? Compare to the speed of light. (d) Assume that the neutron star is uniform and calculate its density. Compare to the density of ordinary rock \(\left(3000 \mathrm{~kg} / \mathrm{m}^{3}\right)\) and to the density of an atomic nucleus (about \(\left.10^{17} \mathrm{~kg} / \mathrm{m}^{3}\right) .\) Justify the statement that a neutron star is essentially a large atomic nucleus.

A uniform bar has two small balls glued to its ends. The bar is \(2.00 \mathrm{~m}\) long and has mass \(4.00 \mathrm{~kg},\) while the balls each have mass \(0.300 \mathrm{~kg}\) and can be treated as point masses. Find the moment of inertia of this combination about an axis (a) perpendicular to the bar through its center; (b) perpendicular to the bar through one of the balls; (c) parallel to the bar through both balls; and (d) parallel to the bar and \(0.500 \mathrm{~m}\) from it.

A flywheel with radius \(0.300 \mathrm{~m}\) starts from rest and accelerates with a constant angular acceleration of \(0.600 \mathrm{rad} / \mathrm{s}^{2} .\) For a point on the rim of the flywheel, what are the magnitudes of the tangential, radial, and resultant accelerations after \(2.00 \mathrm{~s}\) of acceleration?

A fan blade rotates with angular velocity given by \(\omega_{z}(t)=\gamma-\beta t^{2}, \quad\) where \(\quad \gamma=5.00 \mathrm{rad} / \mathrm{s} \quad\) and \(\quad \beta=0.800 \mathrm{rad} / \mathrm{s}^{3}\) (a) Calculate the angular acceleration as a function of time. (b) Calculate the instantaneous angular acceleration \(\alpha_{z}\) at \(t=3.00 \mathrm{~s}\) and the average angular acceleration \(\alpha_{\mathrm{av}-z}\) for the time interval \(t=0\) to \(t=3.00 \mathrm{~s}\). How do these two quantities compare? If they are different, why?
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