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You need to design an industrial turntable that is 60.0 cm in diameter and has a kinetic energy of 0.250 J when turning at 45.0 rpm 1rev/min2. (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
The moment of inertia is approximately 0.0225 kg·m², and the mass is approximately 0.50 kg.

Step by step solution

01

Understand the Problem

The exercise involves calculating the moment of inertia and mass of a turntable designed as a uniform solid disk. First, we know that the diameter is 60.0 cm, implying a radius of 30.0 cm (0.30 m). The turntable has a kinetic energy of 0.250 J and spins at 45.0 rpm.
02

Convert Angular Velocity

Convert the given angular velocity from revolutions per minute (rpm) to radians per second (rad/s). Use the conversion factor \(1 \text{ rev} = 2\pi \text{ rad} \) and \(1 \text{ min} = 60 \text{ s} \).\[\omega = 45.0 \text{ rpm} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} = \frac{45 \times 2\pi}{60} \approx 4.712 \text{ rad/s}\]
03

Relate Kinetic Energy to Moment of Inertia

The rotational kinetic energy \( K \) is related to the moment of inertia \( I \) and angular velocity \( \omega \) by the formula \( K = \frac{1}{2} I \omega^2 \). Rearrange this to solve for the moment of inertia.\[I = \frac{2K}{\omega^2}\]Substitute \( K = 0.250 \text{ J} \) and \( \omega = 4.712 \text{ rad/s} \).\[I = \frac{2 \times 0.250}{4.712^2} \approx 0.0225 \text{ kg} \cdot \text{m}^2\]
04

Use Moment of Inertia Formula for Disk

For a uniform solid disk, the moment of inertia is given by the formula \( I = \frac{1}{2} m r^2 \). Rearrange this to solve for the mass \( m \).\[m = \frac{2I}{r^2}\]Substitute \( I = 0.0225 \text{ kg} \cdot \text{m}^2 \) and \( r = 0.30 \text{ m} \).\[m = \frac{2 \times 0.0225}{0.30^2} \approx 0.50 \text{ kg}\]
05

Validate the Solution

The results are consistent with the physical scenario. A turntable with a diameter of 60.0 cm and described kinetic energy and rotational speed could have a moment of inertia and mass as calculated. Double-check arithmetic in each step to ensure accuracy.

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Key Concepts

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

Rotational Kinetic Energy
Rotational kinetic energy is a type of energy an object possesses due to its rotation. Just like linear kinetic energy depends on mass and velocity, rotational kinetic energy depends on the moment of inertia and angular velocity of an object. For any rotating object, the rotational kinetic energy can be calculated using the formula:
  • \( K = \frac{1}{2} I \omega^2 \)
where \( K \) is the rotational kinetic energy, \( I \) is the moment of inertia, and \( \omega \) is the angular velocity.
To find the rotational kinetic energy, you need to know the object's moment of inertia and its rate of rotation. In the exercise given, the turntable has a kinetic energy of 0.250 J while rotating. This helps us relate the moment of inertia directly to the kinetic energy and the given angular velocity through the equation.
Understanding this relationship is crucial when designing rotating objects such as shafts, wheels, and turntables, as it helps determine how much energy is needed for desired rotational speeds.
Uniform Solid Disk
A uniform solid disk is a simple yet important shape in physics, particularly in rotational dynamics. It represents a two-dimensional object with uniform mass distribution across its volume.
For a uniform solid disk, the calculation of its moment of inertia becomes simplified. This is because such a disk's mass is not concentrated in any one area but is spread evenly throughout the entire disk.
The formula for the moment of inertia of a solid disk about an axis perpendicular to its center is given by:
  • \( I = \frac{1}{2} m r^2 \)
where \( I \) is the moment of inertia, \( m \) is the mass of the disk, and \( r \) is the radius.
Using this formula allows engineers and physicists to easily compute how the mass of the disk affects its rotational characteristics, important for dynamic systems analysis.
Angular Velocity
Angular velocity refers to the rate at which an object rotates or revolves relative to another point, typically the center of a circle it describes. It is an essential concept in rotational dynamics and is a vector quantity, meaning it has both magnitude and direction.
In practice, angular velocity is often given in revolutions per minute (rpm), but for calculations, it is usually converted to radians per second (rad/s). This conversion is done using the conversion factors:
  • \(1 \text{ rev} = 2\pi \text{ rad} \)
  • \(1 \text{ min} = 60 \text{ s} \)
For instance, in the given exercise, the turntable's rotation speed of 45.0 rpm was converted to approximately 4.712 rad/s.
Understanding and calculating angular velocity is vital because it directly influences the rotational kinetic energy an object will have, and is necessary for determining the dynamics of rotating systems.
Mass Calculation
Calculating the mass of an object is a fundamental step in many physics problems, especially when it relates to rotation and motion. In the context of rotational dynamics, knowing the mass helps in determining the object's moment of inertia, which is crucial for understanding its rotational motion.
In the case of a uniform solid disk, once you know the moment of inertia, you can rearrange its formula to find the mass:
  • \(m = \frac{2I}{r^2} \)
where \( I \) is the moment of inertia and \( r \) is the radius.
In the exercise, by knowing the turntable's moment of inertia and its radius (0.30 m), we can calculate its mass, finding it to be approximately 0.50 kg.
This calculation not only aids in understanding the object's dynamics but also ensures that the right material and stability are maintained in the physical world, which is essential for designing functional and safe equipment.

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Most popular questions from this chapter

An electric turntable \(0.750 \mathrm{~m}\) in diameter is rotating about a fixed axis with an initial angular velocity of \(0.250 \mathrm{rev} / \mathrm{s}\) and a constant angular acceleration of \(0.900 \mathrm{rev} / \mathrm{s}^{2}\). (a) Compute the angular velocity of the turntable after \(0.200 \mathrm{~s}\). (b) Through how many revolutions has the turntable spun in this time interval? (c) What is the tangential speed of a point on the rim of the turntable at \(t=0.200 \mathrm{~s} ?\) (d) What is the magnitude of the resultant acceleration of a point on the rim at \(t=0.200 \mathrm{~s} ?\)

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