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Chapter 9: Problem 78

A uniform, solid disk with mass \(m\) and radius \(R\) is pivoted about a horizontal axis through its center. A small object of the same mass \(m\) is glued to the rim of the disk. If the disk is released from rest with the small object at the end of a horizontal radius, find the angular speed when the small object is directly below the axis.

Short Answer

Expert verified
The angular speed \( \omega = \sqrt{\frac{4g}{3R}} \).

Step by step solution

01

Understanding the Problem

We have a solid disk and a small object attached to its rim. Both have the same mass. The disk is pivoted at its center, and we need to find the angular speed when the small object reaches the lowest point below the pivot.
02

Formulating the Conservation of Energy

The potential energy lost by the small object turning from the horizontal position to the bottom position is converted to kinetic energy of the system. The total mechanical energy is conserved.
03

Calculating Initial Potential Energy

Initially, the potential energy of the small object is given by \[ PE_i = mgR \]where \( m \) is its mass, \( g \) is the acceleration due to gravity, and \( R \) is the radius of the disk.
04

Estimating Final Kinetic Energy

At the lowest point, all initial potential energy has converted into rotational kinetic energy of the disk and translational kinetic energy of the small mass. The final kinetic energy is \[ KE_f = \frac{1}{2}I\omega^2 + \frac{1}{2}mv^2 \]Note that \( v = \omega R \). The moment of inertia \( I \) for the disk is \( \frac{1}{2}mR^2 \) and for the point mass \( mR^2 \).
05

Setting Energy Conservation Equation

Equating the initial potential energy to the final kinetic energy:\[ m g R = \frac{1}{2}(\frac{1}{2}mR^2 + mR^2)\omega^2 + \frac{1}{2}(m)(\omega R)^2 \]
06

Solving for Angular Speed

Simplify and solve the above equation for \( \omega \): \[ m g R = \frac{1}{2} \cdot \frac{3}{2}mR^2 \omega^2 \] This simplifies to \[ \omega^2 = \frac{4g}{3R} \] So, \( \omega = \sqrt{\frac{4g}{3R}} \).

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Key Concepts

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

Moment of Inertia
In rotational motion, the moment of inertia is the rotational equivalent of mass in linear motion. Think of it as a measure of an object's resistance to changes in its rotation. For our solid disk, the moment of inertia is given by \[I_{ ext{disk}} = \frac{1}{2}mR^2\]where \( m \) is mass and \( R \) is the radius.
The small point mass glued to the rim also contributes its moment of inertia:\[I_{ ext{mass}} = mR^2\]Combining these gives the total moment of inertia:\[I_{ ext{total}} = \frac{1}{2}mR^2 + mR^2 \]This indicates how both the disk's distribution of mass and the point mass's position affect the system's rotation.
Conservation of Energy
The core concept at play here is the conservation of energy which states that energy in a closed system remains constant over time. It can transform from one form to another, like potential energy becoming kinetic energy, but the total stays the same.
In our problem, the initial potential energy of the small mass at the rim becomes the kinetic energy when it's at the lowest point.
  • Initial potential energy: \( PE_i = mgR \)
  • Final kinetic energy includes both rotational and translational parts.
Angular Speed
Angular speed, represented by \( \omega \), refers to how quickly an object rotates around an axis. It's akin to linear speed in straight-line motion. In our scenario, we wanted to find the angular speed of the disk and the object glued to its rim at the lowest point.
To determine this, we used the conservation of energy principles. By equating the initial potential energy to the total kinetic energy at the bottom, we solved for \( \omega \).
We found:\[\omega = \sqrt{\frac{4g}{3R}}\]showing how the radius of the disk and gravitational acceleration affect rotational speed.
Kinetic Energy
Kinetic energy in rotational dynamics consists of rotational kinetic energy and translational kinetic energy.
For the disk, the rotational kinetic energy is given by:\[KE_{ ext{rotational}} = \frac{1}{2}I \omega^2\]For the small object, its translational kinetic energy is:\[KE_{ ext{translational}} = \frac{1}{2}m(\omega R)^2\]Together, they represent the total energy in motion when the object is at the lowest point.
This conversion from potential to kinetic energy exemplifies how energy shifts forms during rotational movement.
Potential Energy
Potential energy in this scenario is initially stored due to the height of the small mass above the pivot axis.
- The formula for gravitational potential energy is:\[PE = mgh\]Here, \( h = R \) as the initial position is at radius height.- Thus, the initial potential energy is:\[PE_i = mgR\]This energy is vital, as it transforms into kinetic energy, showcasing the energy conversion process during motion, central to solving this rotational problem.

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Most popular questions from this chapter

Centrifuge. An advertisement claims that a centrifuge takes up only 0.127 m of bench space but can produce a radial acceleration of 3000\(g\) at 5000 rev/min. Calculate the required radius of the centrifuge. Is the claim realistic?

A uniform disk with radius \(R=0.400 \mathrm{m}\) and mass 30.0 kg rotates in a horizontal plane on a frictionless vertical axle that passes through the center of the disk. The angle through which the disk has turned varies with time according to \(\theta(t)=\) \((1.10 \mathrm{rad} / \mathrm{s}) t+\left(8.60 \mathrm{rad} / \mathrm{s}^{2}\right) t^{2} .\) What is the resultant linear acceleration of a point on the rim of the disk at the instant when the disk has turned through 0.100 \(\mathrm{rev}\) ?

The spin cycles of a washing machine have two angular speeds, 423 rev/min and 640 rev/min. The internal diameter of the drum is 0.470 \(\mathrm{m}\) . (a) What is the ratio of the maximum radial force on the laundry for the higher angular speed to that for the lower speed? (b) What is the ratio of the maximum tangential speed of the laundry for the higher angular speed to that for the lower speed? (c) Find the laundry's maximum tangential speed and the maximum radial acceleration, in terms of \(g\) .

An airplane propeller is 2.08 \(\mathrm{m}\) in length (from tip to tip) with mass 117 \(\mathrm{kg}\) and is rotating at 2400 \(\mathrm{rpm}(\mathrm{rev} / \mathrm{min})\) about an axis through its center. You can model the propeller as a slender rod. (a) What is its rotational kinetic energy? (b) Suppose that, due to weight constraints, you had to reduce the propeller's mass to 75.0\(\%\) of its original mass, but you still needed to keep the same size and kinetic energy. What would its angular speed have to be, in rpm?

A meter stick with a mass of 0.180 \(\mathrm{kg}\) is pivoted about one end so it can rotate without friction about a horizontal axis. The meter stick is held in a horizontal position and released. As it swings through the vertical, calculate (a) the change in gravitational potential energy that has occurred; (b) the angular speed of the stick; (c) the linear speed of the end of the stick opposite the axis. (d) Compare the answer in part (c) to the speed of a particle that has fallen \(1.00 \mathrm{m},\) starting from rest.
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