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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 \(\mathrm{rpm}(\mathrm{rev} / \mathrm{min}) .\) (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
Moment of inertia is 0.0225 kg·m² and mass is 0.5 kg.

Step by step solution

01

Calculate Angular Velocity

First, convert the rotational speed from revolutions per minute (rpm) to radians per second (rad/s). We use the conversion formulas: \(1 \text{ rpm} = \frac{2\pi}{60} \text{ rad/s}\). For 45.0 rpm, the angular velocity \(\omega\) is calculated as follows:\[ \omega = 45.0 \times \frac{2\pi}{60} \approx 4.71 \text{ rad/s} \]
02

Use the Kinetic Energy Formula

The kinetic energy \(K\) of a rotating object is related to its moment of inertia \(I\) and angular velocity \(\omega\) by the formula:\[ K = \frac{1}{2}I\omega^2 \]Given \(K = 0.250 \text{ J}\) and \(\omega \approx 4.71 \text{ rad/s}\), solve for \(I\):\[ 0.250 = \frac{1}{2}I(4.71)^2 \]\[ I = \frac{0.250 \times 2}{(4.71)^2} \approx 0.0225 \text{ kg} \cdot \text{m}^2 \]
03

Radius of the Turntable

The problem states that the diameter of the turntable is 60.0 cm, which is 0.6 m in SI units. Therefore, the radius \(r\) is:\[ r = \frac{0.6}{2} = 0.3 \text{ m} \]
04

Calculate Mass of the Turntable

If the turntable is a uniform solid disk, its moment of inertia \(I\) is given by the formula:\[ I = \frac{1}{2} m r^2 \]We already have \(I = 0.0225 \text{ kg} \cdot \text{m}^2\) and \(r = 0.3 \text{ m}\). Substitute these into the formula to solve for \(m\):\[ 0.0225 = \frac{1}{2} m (0.3)^2 \]Solving for \(m\):\[ 0.0225 = \frac{1}{2} m (0.09) \]\[ m = \frac{0.0225}{0.045} \approx 0.5 \text{ kg} \]

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Key Concepts

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

Understanding Kinetic Energy of Rotational Motion
Kinetic energy is a vital concept in physics, describing the energy an object possesses due to its motion. When dealing with rotational motion, particularly in objects like a turntable, this energy is referred to as rotational kinetic energy. Similar to linear kinetic energy, which is defined as \( KE = \frac{1}{2}m v^2 \), rotational kinetic energy is given by the formula:\[ K = \frac{1}{2} I \omega^2 \]Where:
  • \( K \) is the kinetic energy
  • \( I \) is the moment of inertia
  • \( \omega \) is the angular velocity
For a turntable spinning at a certain speed, its moment of inertia and angular velocity define its kinetic energy. This relationship highlights how the mass distribution of an object (quantified by \( I \)) and its speed (given by \( \omega \)) contribute to the energy it carries in rotation. Understanding this is crucial in designing systems where energy efficiency and movement precision are a priority.
Calculating the kinetic energy helps engineers ensure that mechanisms like industrial turntables operate within desired energy parameters, balancing speed and safety. For instance, in our exercise, the goal was to achieve a kinetic energy of 0.250 J at a given speed, making these calculations an essential step in the design process.
Decoding Angular Velocity for Rotational Systems
Angular velocity is the rate at which an object rotates around a specific axis. It is analogous to linear velocity in translational motion, but instead of meters per second, it is usually expressed in radians per second (rad/s). Understanding how to convert between different units, such as revolutions per minute (rpm) to rad/s, is crucial for solving problems in rotational dynamics.
In our exercise, the turntable's angular velocity was calculated from 45.0 rpm using the conversion:\[ \omega = 45.0 \times \frac{2\pi}{60} \approx 4.71 \text{ rad/s} \]This conversion is foundational as it translates real-world rotational rates into a format usable in physics equations. It also demonstrates the significance of angular velocity in determining the rotational kinetic energy of an object. The higher the angular velocity, the more kinetic energy an object possesses if the moment of inertia is constant. Understanding and calculating angular velocity can greatly affect the performance and design specifications of rotating machinery. This concept is not just academic; it's applied in various engineering fields including automotive, aerospace, and industrial design to optimize the function and efficiency of rotating components.
The Significance of a Uniform Solid Disk in Moment of Inertia Calculations
A uniform solid disk is a common object in physics problems when examining rotational dynamics. Its symmetrical structure allows for straightforward calculations of its moment of inertia, which is crucial in analyzing its rotational characteristics. The moment of inertia quantifies how much torque is needed for a desired angular acceleration about an axis. For a uniform solid disk, the formula is:\[ I = \frac{1}{2} m r^2 \]Here:
  • \( I \) is the moment of inertia
  • \( m \) is the mass of the disk
  • \( r \) is the radius
The simplicity of this formula helps streamline the design and analysis of rotating systems. In the given example, a calculated moment of inertia allows us to determine how mass should be distributed and the disk's overall mass needed to achieve a specific rotational energy or speed. This property makes the uniform solid disk an ideal choice for applications like industrial turntables where controlled, predictable rotational motion is desired. Moreover, understanding these concepts empowers engineers to design efficient rotating machinery that meets safety and functional specifications while maximizing performance. This ensures that devices are not only effective but resource-efficient.

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Most popular questions from this chapter

A wheel changes its angular velocity with a constant angular acceleration while rotating about a fixed axis through its center. (a) Show that the change in the magnitude of the radial acceleration during any time interval of a point on the wheel is twice the product of the angular acceleration, the angular displacement, and the perpendicular distance of the point from the axis. (b) The radial acceleration of a point on the wheel that is 0.250 \(\mathrm{m}\) from the axis changes from 25.0 \(\mathrm{m} / \mathrm{s}^{2}\) to 85.0 \(\mathrm{m} / \mathrm{s}^{2}\) as the wheel rotates through 20.0 rad. Calculate the tangential acceleration of this point. (c) Show that the change in the wheel's kinetic energy during any time interval is the product of the moment of inertia about the axis, the angular acceleration, and the angular displacement. (d) During the 20.0 -rad angular displacement of part (b), the kinetic energy of the wheel increases from 20.0 \(\mathrm{J}\) to 45.0 \(\mathrm{J} .\) What is the moment of inertia of the wheel about the rotation axis?

A thin, light wire is wrapped around the rim of a wheel, as shown in Fig. E9.49. The wheel rotates without friction about a stationary horizontal axis that passes through the center of the wheel. The wheel is a uniform disk with radius \(R=0.280 \mathrm{m}\) . An object of mass \(m=4.20 \mathrm{kg}\) is suspended from the free end of the wire. The system is released from rest and the suspended object descends with constant acceleration. If the suspended object moves downward a distance of 3.00 \(\mathrm{m}\) in \(2.00 \mathrm{s},\) what is the mass of the wheel?

A thin, rectangular sheet of metal has mass \(M\) and sides of length \(a\) and \(b .\) Use the parallel-axis theorem to calculate the moment of inertia of the sheet for an axis that is perpendicular to the plane of the sheet and that passes through one corner of the sheet.

A disk of radius 25.0 \(\mathrm{cm}\) cm is free to turn about an axle perpendicular to it through its center. It has very thin but strong string wrapped around its rim, and the string is attached to a ball that is pulled tangentially away from the rim of the disk (Fig. P9.65). The pull increases in magnitude and produces an acceleration of the ball that obeys the equation \(a(t)=A t,\) where \(t\) is in seconds and \(A\) is a constant. The cylinder starts from rest, and at the end of the third second, the ball's acceleration is 1.80 \(\mathrm{m} / \mathrm{s}^{2}\) . (a) Find \(A .\) (b) Express the angular acceleration of the disk as a function of time. (c) How much time after the disk has begun to turn does it reach an angular speed of 15.0 \(\mathrm{rad} / \mathrm{s} ?\) (d) Through what angle has the disk turned just as it reaches 15.0 \(\mathrm{rad} / \mathrm{s} ?\) (Hint: See Section \(2.6 .\) .

A bicycle wheel has an initial angular velocity of 1.50 \(\mathrm{rad} / \mathrm{s}\) . (a) If its angular acceleration is constant and equal to 0.300 \(\mathrm{rad} / \mathrm{s}^{2}\) , what is its angular velocity at \(t=2.50 \mathrm{s} ?\) (b) Through what angle has the wheel turned between \(t=0\) and \(t=2.50 \mathrm{s} ?\)

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