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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

As an intern at an engineering firm, you are asked to measure the moment of inertia of a large wheel for rotation about an axis perpendicular to the wheel at its center. You measure the diameter of the wheel to be 0.640 m. Then you mount the wheel on frictionless bearings on a horizontal frictionless axle at the center of the wheel. You wrap a light rope around the wheel and hang an 8.20-kg block of wood from the free end of the rope, as in Fig. E9.45. You release the system from rest and find that the block descends 12.0 m in 4.00 s. What is the moment of inertia of the wheel for this axis?

The angular velocity of a flywheel obeys the equation \(\omega_z\)(\(t\)) \(= A + Bt^2\), where \(t\) is in seconds and \(A\) and \(B\) are constants having numerical values 2.75 (for \(A\)) and 1.50 (for \(B\)). (a) What are the units of \(A\) and \(B\) if \(\omega_z\) is in rad/s? (b) What is the angular acceleration of the wheel at (i) \(t = 0\) and (ii) \(t =\) 5.00 s? (c) Through what angle does the flywheel turn during the first 2.00 s? (\(Hint\): See Section 2.6.)

A bicycle wheel has an initial angular velocity of 1.50 rad/s. (a) If its angular acceleration is constant and equal to 0.200 rad/s\(^2\), what is its angular velocity at \(t =\) 2.50 s? (b) Through what angle has the wheel turned between \(t =\) 0 and \(t =\) 2.50 s?

The angle \(\theta\) through which a disk drive turns is given by \(\theta(t) = a + bt - ct^3\), where \(a\), \(b\), and \(c\) are constants, \(t\) is in seconds, and \(\theta\) is in radians. When \(t =\) 0, \(\theta =\) \(\pi\)/4 rad and the angular velocity is 2.00 rad/s. When \(t =\) 1.50 s, the angular acceleration is 1.25 rad/s\(^2\). (a) Find \(a\), \(b\), and \(c\), including their units. (b) What is the angular acceleration when \(\theta =\) \(\pi\)/4 rad? (c) What are u and the angular velocity when the angular acceleration is 3.50 rad/s\(^2\)?

Two metal disks, one with radius \(R_1 =\) 2.50 cm and mass \(M_1 =\) 0.80 kg and the other with radius \(R_2 =\) 5.00 cm and mass \(M_2 =\) 1.60 kg, are welded together and mounted on a frictionless axis through their common center (\(\textbf{Fig. P9.77}\)). (a) What is the total moment of inertia of the two disks? (b) A light string is wrapped around the edge of the smaller disk, and a 1.50-kg block is suspended from the free end of the string. If the block is released from rest at a distance of 2.00 m above the floor, what is its speed just before it strikes the floor? (c) Repeat part (b), this time with the string wrapped around the edge of the larger disk. In which case is the final speed of the block greater? Explain.

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