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Chapter 11: Problem 12

(II) A potter's whecl is rotating around a vertical axis through its center at a frequency of 1.5 \(\mathrm{rev} / \mathrm{s}\) . The whecl can be considered a uniform disk of mass 5.0 \(\mathrm{kg}\) and diameter 0.40 \(\mathrm{m}\) . The potter then throws a 2.6 -kg chunk of clay, approximately shaped as a flat disk of radius \(8.0 \mathrm{cm},\) onto the center of the rotating whecl. What is the frequency of the whecl after the clay sticks to it?

Short Answer

Expert verified
The new frequency of the wheel is approximately 1.38 rev/s.

Step by step solution

01

Understand the Problem

We need to determine the new frequency of a potter's wheel after a clay chunk sticks to it. This involves the principle of conservation of angular momentum, since no external torques are acting on the system.
02

Define Angular Momentum Conservation

The total angular momentum before the clay is added must equal the total angular momentum after. This can be expressed as: \[ I_1 \omega_1 = (I_1 + I_2) \omega_2 \]where \( I_1 \) and \( \omega_1 \) are the initial moment of inertia and angular velocity, and \( I_2 \) and \( \omega_2 \) are the additional moment of inertia and new angular velocity (frequency) after the clay is added.
03

Calculate the Initial Moment of Inertia \( I_1 \)

For a uniform disk, the moment of inertia is: \[ I_1 = \frac{1}{2} m r^2 \]where \( m = 5.0 \ kg \) (mass of the wheel) and \( r = \frac{0.40}{2} \ m \) (radius of the wheel). \[ I_1 = \frac{1}{2} \times 5.0 \times (0.20)^2 = 0.10 \ \text{kg} \cdot \text{m}^2 \]
04

Calculate Initial Angular Velocity \( \omega_1 \)

The angular velocity in radians per second is given by the formula \[ \omega_1 = 2\pi f \]where \( f \) is the frequency.\[ \omega_1 = 2\pi \times 1.5 \approx 9.42 \ \text{rad/s} \]
05

Calculate Additional Moment of Inertia \( I_2 \)

For the clay disk, which is a flat disk of radius \(0.08 \ m\) and mass \(2.6 \ kg\): \[ I_2 = \frac{1}{2} m r^2 = \frac{1}{2} \times 2.6 \times (0.08)^2 = 0.00832 \ \text{kg} \cdot \text{m}^2 \]
06

Solve for New Angular Velocity \( \omega_2 \)

Using the conservation of angular momentum, we solve for \( \omega_2 \):\[ 0.10 \times 9.42 = (0.10 + 0.00832) \times \omega_2 \]\[ \omega_2 = \frac{0.10 \times 9.42}{0.10832} \approx 8.69 \ \text{rad/s} \]
07

Calculate the New Frequency \( f_2 \)

The new frequency is calculated from the angular velocity:\[ f_2 = \frac{\omega_2}{2\pi} \approx \frac{8.69}{2\pi} \approx 1.38 \ \text{rev/s} \]

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Key Concepts

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

Moment of Inertia
In physics, the moment of inertia is a measure of an object's resistance to changes in its rotation. Think of it like mass for rotation; just as heavier objects are harder to move linearly, objects with larger moments of inertia are harder to spin. It depends on the mass of the object and how this mass is distributed relative to the axis of rotation.
For a uniform disk, which is a symmetrical shape, the formula to calculate the moment of inertia is:
  • \( I = \frac{1}{2}mr^2 \)
where \( m \) is the mass and \( r \) is the radius of the disk. This formula tells us that if all other things are equal, the farther a mass is from the axis of rotation, the larger the moment of inertia will be. Keeping this formula in mind can help you understand why changes in an object's shape or distribution of mass impact its rotational dynamics.
Angular Velocity
Angular velocity represents the rate of rotation and is typically measured in radians per second. It鈥檚 a crucial part of understanding dynamics involving rotating systems. In simple terms, it's how fast something rotates around a particular axis.
To find angular velocity, we often use the relationship:
  • \( \omega = 2\pi f \)
where \( \omega \) is the angular velocity in radians per second, and \( f \) is the frequency in revolutions per second. This relationship captures how frequency and angular velocity are linked through the constant \( 2\pi \), since one revolution corresponds to \( 2\pi \) radians. This concept helps in scenarios like the potter鈥檚 wheel, where knowing the initial frequency allows us to calculate the initial angular momentum and consequently understand how adding mass affects rotational speed.
Uniform Disk
The concept of a uniform disk is important because it simplifies many problems in rotational dynamics. A uniform disk means that its mass is evenly distributed across its shape. This uniformity makes it possible to use simplified equations, as the mass contributing to rotation is spread evenly.
In practical terms:
  • This assumption helps provide a consistent moment of inertia formula: \( I = \frac{1}{2}mr^2 \)
  • The even mass distribution implies that calculations will be uniform and predictable across the disk's area.
This makes a uniform disk a handy model for understanding problems like the potter鈥檚 wheel鈥攐r any situation where objects rotate around a central axis. By representing the wheel and the clay as uniform disks, we can predict how their interaction alters rotational speed using simple physics principles.

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Most popular questions from this chapter

A particle of mass \(m\) uniformly accelerates as it moves counterclockwise along the circumference of a circle of radius \(R\) : $$ \overrightarrow{\mathbf{r}}=\hat{\mathbf{i}} R \cos \theta+\hat{\mathbf{j}} R \sin \theta $$ with \(\theta=\omega_{0} t+\frac{1}{2} \alpha t^{2},\) where the constants \(\omega_{0}\) and \(\alpha\) are the initial angular velocity and angular acceleration, respectively. Determine the object's tangential acceleration \(\overrightarrow{\mathbf{a}}_{\tan }\) and determine the torque acting on the object using \((a) \vec{\tau}=\overrightarrow{\mathbf{r}} \times \overrightarrow{\mathbf{F}}\) (b) \(\vec{\tau}=I \vec{\alpha}\)

A uniform stick \(1.0 \mathrm{~m}\) long with a total mass of \(270 \mathrm{~g}\) is pivoted at its center. A 3.0 -g bullet is shot through the stick midway between the pivot and one end (Fig. \(11-36\) ). The bullet approaches at \(250 \mathrm{~m} / \mathrm{s}\) and leaves at \(140 \mathrm{~m} / \mathrm{s}\). With what angular speed is the stick spinning after the collision?

Water drives a waterwheel (or turbine) of radius \(R=3.0 \mathrm{~m}\) as shown in Fig. \(11-47 .\) The water enters at a speed \(v_{1}=7.0 \mathrm{~m} / \mathrm{s}\) and exits from the waterwheel at a speed \(v_{2}=3.8 \mathrm{~m} / \mathrm{s} . \quad(a)\) If \(85 \mathrm{~kg}\) of water passes through per second, what is the rate at which the water delivers angular momentum to the waterwheel? \((b)\) What is the torque the water applies to the waterwheel? \((c)\) If the water causes the waterwheel to make one revolution every \(5.5 \mathrm{~s}\), how much power is delivered to the wheel?

(1I) A 4.2 -m-diameter merry-go-round is rotating frecly with an angular velocity of 0.80 rad/s. Its total moment of inertia is 1760 \(\mathrm{kg} \cdot \mathrm{m}^{2} .\) Four people standing on the ground, cach of mass 65 \(\mathrm{kg}\) suddenly step onto the edge of the merry-go-round.What is the angular velocity of the merry-go-round now? What if the people were on it initially and then jumped off in a radial direction (relative to the merry-go-round)?

(II) A figure skater can increase her spin rotation rate from an initial rate of 1.0 rev every 1.5 \(\mathrm{s}\) a final rate of 2.5 \(\mathrm{rev} / \mathrm{s} .\) If her initial moment of inertia was 4.6 \(\mathrm{kg} \cdot \mathrm{m}^{2}\) , what is her final moment of inertia? How does she physi- cally accomplish this change?
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