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Chapter 10: Problem 28

A disk like the lower one in Fig. \(10-11\) has a moment of inertia \(I=0.0150 \mathrm{~kg} \cdot \mathrm{m}^{2}\), and is turning at \(3.0 \mathrm{rev} / \mathrm{s}\). A trickle of sand falls onto the revolving disk at a distance of \(20 \mathrm{~cm}\) from the axis and builds a 20 -cm radius narrow ring of sand on it. How much sand must fall on the disk for it to slow to \(2.0 \mathrm{rev} / \mathrm{s} ?\) When a mass \(\Delta m\) of sand falls onto the disk, the moment of inertia of the disk is increased by an amount \(r^{2} \Delta m\), as shown in the preceding problem. After a mass \(m\) has fallen on the disk, the system's moment of inertia has increased to \(I+m r^{2}\). (Note how this agrees with the hoop in Fig.10-1.) Because the sand originally had no angular momentum, the law of conservation of momentum gives $$ \begin{array}{|ll} \text { (Momentum before) }=(\text { Momentum after }) & \text { or } \quad I \omega_{i}=\left(I+m r^{2}\right) \omega_{f} \end{array} $$ from which $$ m=\frac{I\left(\omega_{i}-\omega_{f}\right)}{r^{2} \omega_{f}}=\frac{\left(0.0150 \mathrm{~kg} \cdot \mathrm{m}^{2}\right)(6.0 \pi-4.0 \pi) \mathrm{rad} / \mathrm{s}}{\left(0.040 \mathrm{~m}^{2}\right)(4.0 \pi \mathrm{rad} / \mathrm{s})}=0.19 \mathrm{~kg} $$

Short Answer

Expert verified
The sand must be 0.19 kg.

Step by step solution

01

Understand the Problem

We have a disk with a moment of inertia, initially rotating at a certain speed. Sand falls onto the disk, and we need to find out how much sand is needed to change the disk's rotation speed. We'll use the conservation of angular momentum here.
02

Initial and Final Angular Velocities

Initially, the disk is rotating at \(3.0 \text{ rev/s}\). We convert this to radians per second because angular velocities in equations involving moment of inertia should be in radians per second. Thus, \(\omega_i = 3.0 \times 2\pi = 6\pi \text{ rad/s}\). Finally, the disk slows down to \(2.0 \text{ rev/s}\), so \(\omega_f = 2.0 \times 2\pi = 4\pi \text{ rad/s}\).
03

Conservation of Angular Momentum

Using the law of conservation of angular momentum: \[ I\omega_i = (I + mr^2)\omega_f \]. Here, \(I\) is the initial moment of inertia, \(m\) is the mass of sand that falls on the disk, and \(r\) is the radius where the sand lands.
04

Calculate Mass of Sand

Rearrange the equation to solve for \(m\): \[ m = \frac{I(\omega_i - \omega_f)}{r^2 \omega_f} \]. Given: \(I = 0.0150 \text{ kg}\cdot\text{m}^2\), \(r = 0.20 \text{ m}\), \(\omega_i = 6\pi \text{ rad/s}\), \(\omega_f = 4\pi \text{ rad/s}\).

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Key Concepts

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

Moment of Inertia
Moment of inertia is a central concept in the study of rotational motion, akin to mass in linear motion. It measures how difficult it is to change the rotation of an object. For a rotating body, the moment of inertia depends on both the mass of the body and how that mass is distributed with respect to the axis of rotation.
For example, a solid disk tends to have a different distribution of mass compared to a hoop or a ring, even if they have the same mass. This is why the moment of inertia is also known as the rotational inertia.

In our exercise, we're dealing with a disk that already has a moment of inertia, denoted as \(I = 0.0150 \, \text{kg}\cdot \text{m}^2\). When sand is added to the disk, its moment of inertia changes because the distribution of mass changes. The additional inertia due to the sand is calculated by \(r^2 \Delta m\), where \(r\) is the radius from the axis at which the sand accumulates.

This ability to calculate adjustments to the moment of inertia by simply adding \(mr^2\) when new mass is added, makes problem-solving more straightforward.
Rotational Motion
Rotational motion is movement about an axis. It's the rotational equivalent of linear motion and is described by similar concepts like velocity and acceleration but in terms of angles. Key terms and concepts involved in rotational motion include
  • Angular velocity \(\omega\), which measures how fast something is rotating in radians per second.
  • Angular momentum, which is the rotational analogue to linear momentum, represented as \(I\omega\).

In rotational systems, the conservation laws play a significant role, especially the conservation of angular momentum. This principle states that unless acted upon by an external torque, the total angular momentum remains constant in an isolated system.

In the exercise, the sand being added affects the disk's rotational motion but doesn't add external torque since it falls symmetrically. The law of conservation of angular momentum simplifies the problem, allowing us to equate initial and final angular momentum to solve for unknowns like the amount of sand needed.

Physics Problem Solving
When tackling a physics problem like the one in our exercise, it is important to follow some key steps:
  • Understand the problem: Identify the known and unknown variables. Define what physical concepts can be applied.
  • Draw a diagram or a model: Visualize the system to better understand how forces or changes occur. In our case, a disk with a radius where sand falls helps understand moment of inertia and rotational motion interactions.
  • Use appropriate physics principles: Apply laws like conservation of angular momentum to connect initial conditions with final ones.
  • Solve the equations: Algebraically manipulate the relevant equations to isolate and solve for desired quantities, such as mass in this case.
  • Check units and results: Ensure that your calculations are consistent and reasonable by verifying units and seeing if they match expected physical intuition.

Remember, practice with problem-solving not only helps with solving similar questions but also captures the essence of understanding theories in practical scenarios. Breaking down complex motions into simpler, manageable steps is the goal!

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Most popular questions from this chapter

A \(500-\mathrm{g}\) wheel that has a moment of inertia of \(0.015 \mathrm{~kg} \cdot \mathrm{m}^{2}\) is initially turning at \(30 \mathrm{rev} / \mathrm{s}\). It coasts uniformly to rest after 163 rev. How large is the torque that slowed it?

As a solid disk rolls up and over the top of a hill on a track, its speed slows to \(80 \mathrm{~cm} / \mathrm{s}\). It subsequently descends down the other side of the hill. If friction losses are negligible, how fast is the disk moving when it is \(18 \mathrm{~cm}\) below the top? At the top, the disk has translational and rotational \(\mathrm{KE}\), plus its \(\mathrm{PE}_{\mathrm{G}}\) relative to the point \(18 \mathrm{~cm}\) below. At that final point, \(\mathrm{PE}_{\mathrm{G}}\) has been transformed to more \(\mathrm{KE}\) of rotation and translation. Conservation of energy can be expressed as $$ \begin{array}{l} \left(\mathrm{KE}_{t}+\mathrm{KE}_{r}\right)_{\text {start }}+M g h=\left(\mathrm{KE}_{t}+\mathrm{KE}_{r}\right)_{\text {end }} \\ \frac{1}{2} M v_{i}^{2}+\frac{1}{2} I \omega_{i}^{2}+M g h=\frac{1}{2} M v_{f}^{2}+\frac{1}{2} I \omega_{f}^{2} \end{array} $$ For a solid disk, \(I=\frac{1}{2} M r^{2}\). Also, \(\omega=v / r\). Substituting these values and simplifying yields $$ \frac{1}{2} v_{i}^{2}+\frac{1}{4} v_{i}^{2}+g h=\frac{1}{2} v_{f}^{2}+\frac{1}{4} v_{f}^{2} $$ Employing \(v_{i}=0.80 \mathrm{~m} / \mathrm{s}\) and \(h=0.18 \mathrm{~m}\), substitution gives \(v_{f}=1.7 \mathrm{~m} / \mathrm{s}\).

Suppose that a satellite goes around the Moon in an elliptical orbit. At its closest approach it has a speed \(v_{c}\) and a radius \(r_{c}\) from the center of the Moon. At its farthest distance, it has a speed \(v_{f}\) and a radius \(r_{f}\). Find the ratio \(v_{c} / v_{f}\). [Hint: Angular momentum is conserved, and, moreover, the satellite can be treated as a point mass.]

A 90 -kg person stands at the edge of a stationary children's merry-go-round (essentially a disk) at a distance of \(5.0 \mathrm{~m}\) from its center. The person starts to walk around the perimeter of the disk at a speed of \(0.80 \mathrm{~m} / \mathrm{s}\) relative to the ground. What rotation rate does this motion impart to the disk if \(I_{\text {disk }}=20000 \mathrm{~kg} \cdot \mathrm{m}^{2} ?\left[\right.\) Hint \(:\) For the person, \(\left.I=m r^{2} .\right]\)

A wheel and axle having a total moment of inertia of \(0.0020 \mathrm{~kg} \cdot \mathrm{m}^{2}\) is caused to rotate about a horizontal axis by means of an \(800-\mathrm{g}\) mass attached to a weightless cord wrapped around the axle. The radius of the axle is \(2.0 \mathrm{~cm}\). Starting from rest, how far must the mass fall to give the wheel a rotational rate of \(30 \mathrm{rev} / \mathrm{s} ?\)
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