/*! 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 38 A wheel is turning about an axis... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(\mathrm{J} .\) For an axis through its center, what is the moment of inertia of the wheel?

Short Answer

Expert verified
The moment of inertia is approximately 0.125 kg·m².

Step by step solution

01

Convert Revolutions to Radians

Since the wheel turns 8.20 revolutions, we first need to convert this to radians because angular measurements are often required in radians for calculations involving angular motion. 1 revolution = \(2 \pi\) radians. Therefore, 8.20 revolutions = \(8.20 \times 2 \pi = 16.4\pi\) radians.
02

Determine Angular Displacement and Equations of Motion

The angular displacement \( \theta \) is \(16.4\pi\) radians. Since the wheel started from rest and has constant angular acceleration, we use the equation for angular displacement: \[ \theta = \omega_0 t + \frac{1}{2} \alpha t^2, \]where \( \omega_0 = 0 \) (since it starts from rest), \( \alpha \) is the angular acceleration, and \( t = 12.0 \) s. We substitute the given values:\[ 16.4\pi = \frac{1}{2} \alpha (12.0)^2 \]Solving for \( \alpha \), we get:\[ \alpha = \frac{16.4\pi}{72} \].
03

Calculate Angular Velocity at t = 12 s

Now that we know the angular acceleration, we can find the angular velocity \( \omega \) at \( t = 12.0 \) s using:\[ \omega = \omega_0 + \alpha t, \]where \( \omega_0 = 0 \). So:\[ \omega = \frac{16.4\pi}{72} \times 12.0 = \frac{16.4\pi}{6} \].
04

Use Kinetic Energy to Find Moment of Inertia

The kinetic energy \(K\) at \( t = 12 \) s is given as \(36.0\) J. The kinetic energy of a rotating object is given by:\[ K = \frac{1}{2} I \omega^2, \]where \( I \) is the moment of inertia. We have \( K = 36.0 \) J and \( \omega = \frac{16.4\pi}{6} \). Substitute these values into the equation:\[ 36.0 = \frac{1}{2} I \left(\frac{16.4\pi}{6}\right)^2 \].Solve for \( I \):\[ I = \frac{36.0 \times 2 \times 36}{(16.4\pi)^2} \].
05

Simplify to Find the Numerical Value of Moment of Inertia

Calculate \((16.4\pi)^2\) and solve for \( I \):\[ I = \frac{36.0 \times 2 \times 36}{16.4^2 \times \pi^2} \].This simplifies to:\[ I = \frac{2592}{16.4^2 \times \pi^2} \].Perform the calculations to find the numerical value of \( I \).

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Key Concepts

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

Angular Acceleration
Angular acceleration refers to how quickly an object speeds up or slows down its rotation. It's the rotational equivalent of linear acceleration, describing the rate at which the angular velocity changes over time. In our exercise, the wheel starts from rest and begins to accelerate at a constant rate, which means the angular acceleration is uniform.
To calculate angular acceleration, we use the formula which relates angular displacement (\(\theta\)), initial angular velocity (\(\omega_0\)), time (\(t\)), and angular acceleration (\(\alpha\)). The formula is:
  • \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \)
Since the wheel starts from rest, \(\omega_0\) is 0, simplifying our equation to \(\theta = \frac{1}{2} \alpha t^2\). By substituting the given values, you can solve for \(\alpha\), which describes how quickly the wheel's rotation increases.
Angular Velocity
Angular velocity tells us how fast an object is rotating. It is the rate of change of angular displacement over time. Think of it like the speed of a rotating object measured in radians per second.
To find the angular velocity at any point in time, once the angular acceleration is known, we use the formula:
  • \( \omega = \omega_0 + \alpha t \)
In this instance, since the initial angular velocity \(\omega_0\) is zero (the wheel starts from rest), the angular velocity at 12 seconds is simply \( \omega = \alpha t \). This means that the future angular velocity (\(\omega\)) depends on both the initial angular velocity and how long the object has been accelerating.
Kinetic Energy
Kinetic energy in rotating objects pertains to the energy an object possesses due to its motion. For rotary motion, the formula is slightly different from that of linear motion and is given by:
  • \( K = \frac{1}{2} I \omega^2 \)
Here, \(K\) represents the kinetic energy, \(I\) is the moment of inertia, and \(\omega\) is the angular velocity. The moment of inertia is a property that reflects how the mass of an object is distributed relative to the axis about which it rotates. The wheel's kinetic energy at 12 seconds is given, allowing for the computation of the moment of inertia by rearranging the formula. This real-world application of kinetic energy illustrates how mass distribution and angular speed affect the rotational energy of an object.
Radians Conversion
Radians are a unit of angular measure used extensively in rotational motion problems. Unlike degrees, which divide a circle into 360 equal parts, radians consider the circle's circumference as the measure. One full revolution corresponds to \(2\pi\) radians.
Converting from revolutions to radians is fundamental in many physics problems because the equations of motion typically use radians as their measure of angular displacement.
  • For example, to convert 8.20 revolutions into radians: 8.20 revolutions = \(8.20 \times 2\pi = 16.4\pi\) radians.
This conversion ensures that equations related to angular displacement and angular velocity are accurate, as they rely on radians to produce correct results.

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Most popular questions from this chapter

A wheel is rotating about an axis that is in the \(z\) -direction. The angular velocity \(\omega_{z}\) is \(-6.00\) rad/s at \(t=0\) , increases lin early with time, and is \(+8.00 \mathrm{rad} / \mathrm{s}\) at \(t=7.00 \mathrm{s}\) . We have taken counterclockwise rotation to be positive. (a) Is the angular acceleration during this time interval positive or negative? (b) During what time interval is the speed of the wheel increasing? Decreasing? (c) What is the angular displacement of the wheel at \(t=7.00\) s?

While riding a multispeed bicycle, the rider can select the radius of the rear sprocket that is fixed to the rear axle. The front sprocket of a bicycle has radius 12.0 \(\mathrm{cm} .\) If the angular speed of the front sprocket is 0.600 rev/s, what is the radius of the rear sprocket for which the tangential speed of a point on the rim of the rear wheel will be 5.00 \(\mathrm{m} / \mathrm{s} ?\) The rear wheel has radius 0.330 \(\mathrm{m} .\)

A hollow spherical shell has mass 8.20 \(\mathrm{kg}\) and radius 0.220 \(\mathrm{m} .\) It is initially at rest and then rotates about a stationary axis that lies along a diameter with a constant acceleration of 0.890 \(\mathrm{rad} / \mathrm{s}^{2} .\) What is the kinetic energy of the shell after it has turned through 6.00 \(\mathrm{rev} ?\)

A sphere consists of a solid wooden ball of uniform density 800 \(\mathrm{kg} / \mathrm{m}^{3}\) and radius 0.30 \(\mathrm{m}\) and is covered with a thin coating of lead foil with area density 20 \(\mathrm{kg} / \mathrm{m}^{2}\) . Calculate the moment of inertia of this sphere about an axis passing through its center.

You are to design a rotating cylindrical axle to lift \(800-\mathrm{N}\) buckets of cement from the ground to a rooftop 78.0 \(\mathrm{m}\) above the ground. The buckets will be attached to a hook on the free end of a cable that wraps around the rim of the axle; as the axle turns, the buckets will rise. (a) What should the diameter of the axle be in order to raise the buckets at a steady 2.00 \(\mathrm{cm} / \mathrm{s}\) when it is turning at 7.5 \(\mathrm{rpm}\) (b) If instead the axle must give the buckets an upward acceleration of \(0.400 \mathrm{m} / \mathrm{s}^{2},\) what should the angular acceleration of the axle be?

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