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In \(5.00 \mathrm{~s}\), a \(2.00 \mathrm{~kg}\) stone moves in a horizontal circle of radius \(2.00 \mathrm{~m}\) from rest to an angular speed of \(4.00 \mathrm{rev} / \mathrm{s}\). What are the stone's (a) average angular acceleration and (b) rotational inertia around the circle's center?

Short Answer

Expert verified
(a) Average angular acceleration is \(\frac{8\pi}{5} \, \text{rad/s}^2\). (b) Rotational inertia is \(8.00 \, \text{kg} \cdot \text{m}^2\).

Step by step solution

01

Convert angular speed to radians per second

The final angular speed is given as \(4.00 \, \text{rev/s}\). First, convert revolutions to radians: \(1 \, \text{revolution} = 2\pi \, \text{radians}\). Thus, \(4.00 \, \text{rev/s} = 4.00 \times 2\pi = 8\pi \, \text{rad/s}\).
02

Calculate Average Angular Acceleration

Average angular acceleration \(\alpha\) can be calculated using \(\alpha = \frac{\Delta \omega}{\Delta t}\), where \(\Delta \omega\) is the change in angular speed and \(\Delta t\) is the time interval. Here, \(\Delta \omega = 8\pi \, \text{rad/s} - 0 \, \text{rad/s}\), and \(\Delta t = 5.00 \, \text{s}\). Thus, \(\alpha = \frac{8\pi}{5} = \frac{8\pi}{5} \, \text{rad/s}^2\).
03

Calculate Rotational Inertia

The rotational inertia \(I\) for a point mass going in a circle is given by \(I = m \cdot r^2\), where \(m = 2.00 \, \text{kg}\) is the mass and \(r = 2.00 \, \text{m}\) is the radius. Therefore, \(I = 2.00 \times (2.00)^2 = 8.00 \, \text{kg} \cdot \text{m}^2\).

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Key Concepts

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

Angular Acceleration
Angular acceleration is an important concept in rotational motion. It describes how quickly an object speeds up or slows down its rotation. Just as linear acceleration refers to the change in linear velocity over time, angular acceleration is the change in angular speed over time.

To compute the average angular acceleration \( \alpha \) of an object moving in a circle, use the formula:

\[ \alpha = \frac{\Delta \omega}{\Delta t} \]

where:
  • \( \Delta \omega \) is the change in angular speed.
  • \( \Delta t \) is the change in time.
For example, if an object starts from rest and achieves an angular speed of \(8\pi \text{ rad/s}\) in \(5\text{ s}\), the average angular acceleration would be \( \frac{8\pi}{5} \text{ rad/s}^2\).

This calculation shows how angular acceleration helps analyze how rotational speed evolves, providing insights into the forces acting on the object.
Rotational Inertia
Rotational inertia, also known as the moment of inertia, is a key concept in understanding rotation. It measures an object's resistance to changes in its rotational motion.

This is similar to how mass measures resistance to changes in linear motion. In essence, the larger the mass and the further the mass is from the rotation axis, the greater the rotational inertia.

The rotational inertia \( I \) for a point mass moving in a circle is calculated as:

\[ I = m \cdot r^2 \]

where:
  • \( m \) is the mass of the object.
  • \( r \) is the radius of the circle.
For instance, if a \( 2.00 \text{ kg} \) stone swings around a circle with a radius of \( 2.00 \text{ m} \), the rotational inertia would be \( 8.00 \text{ kg} \cdot \text{m}^2 \).

This understanding of rotational inertia explains why objects with more mass further from their axis are harder to spin and stop spinning.
Angular Speed Conversion
Angular speed tells us how fast an object rotates or revolves and is often calculated in revolutions per second (rev/s). However, physics problems typically require angular speed in radians per second (rad/s).

The conversion is straightforward since one complete revolution corresponds to \(2\pi\) radians.

To convert angular speed from rev/s to rad/s, multiply the value by \(2\pi\).

For example, an angular speed of \(4.00 \text{ rev/s}\) can be converted to rad/s by:

\[ 4.00 \times 2\pi = 8\pi \text{ rad/s} \]

This process is crucial for solving physics problems because many equations, like those for angular acceleration and torque, require angular speed in radians per second. By converting units, you ensure your calculations are consistent with standard formulas used in rotational dynamics.

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Most popular questions from this chapter

Disks \(A\) and \(B\) each have a rotational inertia of \(0.300 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about the central axis and a radius of \(20.0 \mathrm{~cm}\) and are free to rotate on a central rod through both of them. To set them spinning around the rod in the same direction, each is wrapped with a string that is then pulled for \(10.0 \mathrm{~s}\) (the string detaches at the end). The magnitudes of the forces pulling the strings are \(30.0 \mathrm{~N}\) for disk \(A\) and \(20.0 \mathrm{~N}\) for disk \(B\). After the strings detach, the disks happen to collide and the frictional force between them brings them to the same final angular speed in \(6.00 \mathrm{~s}\). What are (a) magnitude of the average frictional torque that brings them to the final angular speed and (b) the loss in kinetic energy as that torque acts on them? (c) Where did the "lost energy" go?

A gyroscope flywheel of radius \(2.62 \mathrm{~cm}\) is accelerated from rest at \(14.2 \mathrm{rad} / \mathrm{s}^{2}\) until its angular speed is \(2760 \mathrm{rev} / \mathrm{min}\). (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?

Just as a helicopter is landing, its blades are turning at \(30.0 \mathrm{rev} / \mathrm{s}\) and slowing at a constant rate. In the \(2.00 \mathrm{~min}\) required for them to stop, how many revolutions do they make?

Starting from rest, a disk rotates about its central axis with constant angular acceleration. In \(5.0 \mathrm{~s}\), it rotates \(20 \mathrm{rad}\). 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) A uniform \(2.00 \mathrm{~kg}\) disk of radius \(0.300 \mathrm{~m}\) can rotate around its central axis like a merry-go-round. Beginning from resi at time \(t=0\), it undergoes a constant angular acceleration of \(30.0 \mathrm{rad} / \mathrm{s}^{2}\). When is the rotational kinetic energy equal to \(2000 \mathrm{~J}\) (b) Repeat the calculation but substitute a ring of the same \(\omega(\mathrm{rad} / \mathrm{s})\) substitute a ring of the same mass and radius and assume that the spokes have negligible mass.

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