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A wheel is turning about an axis through its center with constant angular acceleration. Starting from rest, at \(t =\) 0, the wheel turns through 8.20 revolutions in 12.0 s. At \(t =\) 12.0 s the kinetic energy of the wheel is 36.0 J. For an axis through its center, what is the moment of inertia of the wheel?

Short Answer

Expert verified
The moment of inertia of the wheel is approximately 0.979 kg⋅m².

Step by step solution

01

Convert Revolutions to Radians

First, let's convert the revolutions into radians since 1 revolution is equal to \(2\pi\) radians. The wheel turns through 8.20 revolutions, so the angular displacement \(\theta\) is \(8.20 \times 2\pi = 16.40\pi \text{ radians}\).
02

Use Angular Displacement Formula

The formula for angular displacement with constant angular acceleration \(\alpha\) starting from rest is given by:\[\theta = \frac{1}{2} \alpha t^2\]Substituting \(\theta = 16.40\pi\) and \(t = 12.0\, \text{s}\), we solve for \(\alpha\):\[16.40\pi = \frac{1}{2} \alpha (12)^2\]Solving for \(\alpha\) gives \(\alpha = \frac{16.40\pi}{72} = \frac{41\pi}{180} \approx 0.715\, \text{rad/s}^2\).
03

Calculate Final Angular Velocity

Using the formula for final angular velocity \(\omega\) with constant angular acceleration:\[\omega = \omega_0 + \alpha t\]where \(\omega_0 = 0\) (since it starts from rest), we substitute to find:\[\omega = 0 + 0.715 \times 12 = 8.58\, \text{rad/s}\].
04

Use Kinetic Energy Formula to Find Moment of Inertia

The kinetic energy of a rotating object is given by:\[KE = \frac{1}{2} I \omega^2\]where \(KE = 36.0\, \text{J}\) and \(\omega = 8.58\, \text{rad/s}\). Rearrange the equation to solve for the moment of inertia \(I\):\[I = \frac{2 \times 36.0}{8.58^2} = \frac{72.0}{73.5364} \approx 0.979\, \text{kg} \cdot \text{m}^2\].

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Key Concepts

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

Angular Displacement
Angular displacement refers to the angle through which a point, line, or body is rotated in a specified direction and about a specified axis. In rotational mechanics, it is crucial because it represents how much rotation has occurred.
In our example, the wheel turns through 8.20 revolutions, which needs to be translated into radians for most calculations. Since one full revolution equals \(2\pi\) radians, the angular displacement \(\theta\) becomes \(16.40\pi\) radians.
  • Revolutions to radians conversion is essential in these problems.
  • Angular displacement helps understand the extent of rotation.
Understanding angular displacement aids in calculating other rotational quantities like angular velocity and acceleration.
Angular Velocity
Angular velocity describes how fast an object rotates or revolves relative to another point, often the center of rotation. This measure is typically in radians per second, offering insights into the rotational speed of objects.In the exercise, the wheel starts from rest, meaning its initial angular velocity \(\omega_0\) is 0. As it accelerates, its angular velocity increases to \(8.58\, \text{rad/s}\) at \(t = 12\, \text{s}\). This increase results from the constant angular acceleration of \(0.715\, \text{rad/s}^2\):
  • \(\omega = \omega_0 + \alpha t\)
  • For objects starting from rest: \(\omega = \alpha t\)
Angular velocity connects the rotational speed and time, crucial for understanding rotational dynamics.
Angular Acceleration
Angular acceleration indicates the rate of change of angular velocity over time. It fills a similar role in rotational motion as linear acceleration does in linear motion. Measured in radians per second squared, it is a vital factor in calculating how quickly an object speeds up or slows down its rotation.From the exercise, starting from rest, the angular displacement formula \(\theta = \frac{1}{2} \alpha t^2\) was used to find the angular acceleration \(\alpha\). With a solution of approximately \(0.715\, \text{rad/s}^2\), it shows a steady increase of angular velocity over the 12 seconds.
  • This represents constant acceleration in rotational motion.
  • Critical for calculating the subsequent angular velocity.
Angular acceleration is foundational in determining how quickly rotational motion evolves.
Kinetic Energy
In rotational motions, kinetic energy provides insight into the energy that a rotating object possesses due to its motion. Like linear kinetic energy, rotational kinetic energy depends on the motion state and properties of the object.For the wheel in question, kinetic energy relies on the wheel's moment of inertia \(I\) and its angular velocity \(\omega\). The formula used is \(KE = \frac{1}{2} I \omega^2\), where the kinetic energy at \(t = 12.0\, \text{s}\) is given as 36.0 J. By rearranging this equation, the moment of inertia is calculated to be approximately \(0.979\, \text{kg} \cdot \text{m}^2\).
  • Shows how the wheel's ability to resist rotational changes relates to its energy.
  • Relates energy and rotational dynamics.
Understanding kinetic energy in rotational contexts provides a complete picture of an object's motion state and corresponding energy.

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Most popular questions from this chapter

A light, flexible rope is wrapped several times around a \(hollow\) cylinder, with a weight of 40.0 N and a radius of 0.25 m, that rotates without friction about a fixed horizontal axis. The cylinder is attached to the axle by spokes of a negligible moment of inertia. The cylinder is initially at rest. The free end of the rope is pulled with a constant force \(P\) for a distance of 5.00 m, at which point the end of the rope is moving at 6.00 m/s. If the rope does not slip on the cylinder, what is \(P\)?

A sphere with radius \(R =\) 0.200 m has density \(\rho\) that decreases with distance \(r\) from the center of the sphere according to \(r =\) 3.00 \(\times\) 103 kg/m\(^3 -\) (9.00 \(\times\) 103 kg/m\(^4\))\(r\). (a) Calculate the total mass of the sphere. (b) Calculate the moment of inertia of the sphere for an axis along a diameter.

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The rotating blade of a blender turns with constant angular acceleration 1.50 rad/s\(^2\). (a) How much time does it take to reach an angular velocity of 36.0 rad/s, starting from rest? (b) Through how many revolutions does the blade turn in this time interval?

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