/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 Calculate the rotational inertia... [FREE SOLUTION] | 91Ó°ÊÓ

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Calculate the rotational inertia of a wheel that has a kinetic energy of \(24400 \mathrm{~J}\) when rotating at 602 rev/min.

Short Answer

Expert verified
The rotational inertia of the wheel is approximately 12.280 kg·m².

Step by step solution

01

Convert Angular Velocity

First, convert the angular velocity from revolutions per minute to radians per second. Since there are \(2 \pi\) radians in one revolution and 60 seconds in a minute, we have:\[ \omega = 602 \text{ rev/min} \times \frac{2 \pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} = \frac{602 \times 2 \pi}{60} \text{ rad/s} \approx 63.023 \text{ rad/s} \]
02

Use the Rotational Kinetic Energy Formula

The rotational kinetic energy \(K\) of a rotating object is given by the formula:\[ K = \frac{1}{2} I \omega^2 \]where \(K = 24400 \, \text{J}\) and \(\omega = 63.023 \, \text{rad/s}\). We need to solve for the rotational inertia \(I\).
03

Solve for Rotational Inertia

Rearrange the formula for rotational kinetic energy to solve for the rotational inertia \(I\):\[ I = \frac{2K}{\omega^2} \]Substitute the known values:\[ I = \frac{2 \times 24400}{(63.023)^2} \]Compute \(I\):\[ I = \frac{48800}{3973.899} \approx 12.280 \, \text{kg} \cdot \text{m}^2 \]
04

Finalize the Answer

The calculated rotational inertia of the wheel is approximately \(12.280 \, \text{kg} \cdot \text{m}^2\).

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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 form of kinetic energy that is associated with the rotation of an object. It is similar to translational kinetic energy, which is associated with the linear motion of an object. However, instead of dealing with linear velocity, rotational kinetic energy deals with angular velocity. This energy can be defined using the formula:\[ K = \frac{1}{2} I \omega^2 \]where:
  • \( K \) is the rotational kinetic energy in joules (J),
  • \( I \) is the rotational inertia, or moment of inertia, measured in kgâ‹…m²,
  • \( \omega \) is the angular velocity in radians per second (rad/s).
Understanding this concept helps us appreciate how energy is stored in rotating systems, such as wheels, gears, or even celestial objects like planets. Whenever something spins, it possesses rotational kinetic energy, which can be calculated if the rotational inertia and angular velocity are known. Remember, rotational inertia depends on how mass is distributed relative to the axis of rotation, making it a crucial factor in the energy equation.
Angular Velocity Conversion
When dealing with rotational motion, it's common to encounter angular velocity in different units, such as revolutions per minute (rev/min) or radians per second (rad/s). Understanding how to convert between these units is essential for solving problems related to rotational motion. To convert from revolutions per minute to radians per second, remember that:
  • There are \( 2\pi \) radians in one revolution.
  • There are 60 seconds in a minute.
Thus, the conversion formula becomes:\[ \omega = \text{(rev/min)} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} \] This conversion is crucial for using the rotational kinetic energy formula, which requires angular velocity in radians per second. For example, if we have an angular velocity of 602 rev/min, converting it gives us approximately 63.023 rad/s, enabling us to move forward with energy calculations.
Radians per Second
The unit "radians per second" (rad/s) is the standard unit of angular velocity in physics. It quantifies how quickly an object rotates or spins around a particular axis. Unlike units like degrees per second or revolutions per minute, radians per second directly relate to the mathematical properties of circles and rotational motion.One complete revolution equals \( 2\pi \) radians, making the radian a natural measure for rotations. Using radians per second allows for more straightforward mathematical treatments of rotational problems, particularly when calculating rotational kinetic energy or other properties of rotating systems.Starting from a known rotational speed in revolutions per minute or another unit, converting to radians per second ensures precision and consistency in our calculations. This consistency is vital for ensuring our solutions in physics applications remain accurate and reliable.
Kinetic Energy Formula
The kinetic energy formula for rotating objects is a vital tool in physics. It allows us to calculate the energy associated with rotational motion, which is important in understanding dynamics of spinning objects. The formula is written as:\[ K = \frac{1}{2} I \omega^2 \] Where:
  • \( K \) is the kinetic energy in joules,
  • \( I \) is the rotational inertia, indicating the distribution of mass,
  • \( \omega \) is the angular velocity.
To find the rotational inertia \( I \), you rearrange the formula to:\[ I = \frac{2K}{\omega^2} \]This rearranged formula is particularly helpful when you know the energy and angular velocity but need to determine the rotational inertia of the object. Analyzing the kinetic energy in terms of rotation provides insight into the behavior of spinning bodies, such as wheels or propellers, contributing to fields ranging from mechanical engineering to space exploration.

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Most popular questions from this chapter

A golf ball is launched at an angle of \(20^{\circ}\) to the horizontal, with a speed of \(60 \mathrm{~m} / \mathrm{s}\) and a rotation rate of \(90 \mathrm{rad} / \mathrm{s}\). Neglecting air drag, determine the number of revolutions the ball makes by the time it reaches maximum height.

Starting from rest, a disk rotates about its central axis with constant angular acceleration. In \(5.0 \mathrm{~s}\), it rotates \(25 \mathrm{rad}\). During that time, what are the magnitudes of (a) the angular acceleration and (b) the average angular velocity? (c) What is the instantaneous angular velocity of the disk at the end of the \(5.0 \mathrm{~s}\) ? (d) With the angular acceleration unchanged, through what additional angle will the disk turn during the next \(5.0 \mathrm{~s}\) ?

A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is rotating at \(10 \mathrm{rev} / \mathrm{s} ; 60\) revolutions later, its angular speed is 15 rev/s. Calculate (a) the angular acceleration, (b) the time required to complete the 60 revolutions, (c) the time required to reach the \(10 \mathrm{rev} / \mathrm{s}\) angular speed, and (d) the number of revolutions from rest until the time the disk reaches the 10 rev/s angular speed.

A high-wire walker always attempts to keep his center of mass over the wire (or rope). He normally carries a long, heavy pole to help: If he leans, say, to his right (his com moves to the right) and is in danger of rotating around the wire, he moves the pole to his left (its com moves to the left) to slow the rotation and allow himself time to adjust his balance. Assume that the walker has a mass of \(70.0 \mathrm{~kg}\) and a rotational inertia of \(15.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about the wire. What is the magnitude of his angular acceleration about the wire if his com is \(5.0 \mathrm{~cm}\) to the right of the wire and (a) he carries no pole and (b) the \(14.0 \mathrm{~kg}\) pole he carries has its \(\operatorname{com} 10 \mathrm{~cm}\) to the left of the wire?

A gyroscope flywheel of radius \(2.83 \mathrm{~cm}\) is accelerated from rest at \(14.2 \mathrm{rad} / \mathrm{s}^{2}\) until its angular speed is 2760 rev/min. (a) What is the tangential acceleration of a point on the rim of the flywheel during this spin-up process? (b) What is the radial acceleration of this point when the flywheel is spinning at full speed? (c) Through what distance does a point on the rim move during the spin-up?

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