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A \(0.015-\mathrm{kg}\) record with a radius of \(15 \mathrm{cm}\) rotates with an angular speed of \(33 \frac{1}{3} \mathrm{rpm} .\) Find the angular momentum of the record.

Short Answer

Expert verified
The angular momentum is \(0.0005896\,\text{kg}\cdot\text{m}^2/\text{s}.\)

Step by step solution

01

Identify the Given Values

We have a record with a mass of \(m = 0.015\,\text{kg}\), a radius of \(r = 15\,\text{cm}\) which can be converted to \(0.15\,\text{m}\), and an angular speed of \(\omega = 33\,\frac{1}{3}\,\text{rpm}\). First, we need to convert \(\omega\) to \(\text{rad/s}\).
02

Convert RPM to Radians Per Second

To convert \(33\,\frac{1}{3}\,\text{rpm}\) to radians per second: 33\,\frac{1}{3}\,\text{rpm} = \frac{100}{3}\,\text{rpm}. 1\,\text{rpm} = \frac{2\pi}{60}\,\text{rad/s}. \(\omega = \frac{100}{3}\times \frac{2\pi}{60}\,\text{rad/s} = \frac{200\pi}{180}\,\text{rad/s} = \frac{10\pi}{9}\,\text{rad/s}.\)
03

Calculate the Moment of Inertia

Assuming the record is a solid disk, its moment of inertia \(I\) is given by: \(I = \frac{1}{2}mr^2 = \frac{1}{2}(0.015)(0.15)^2\,\text{kg}\cdot\text{m}^2.\) Solve for \(I\): \(I = \frac{1}{2}(0.015)(0.0225) = 0.00016875\,\text{kg}\cdot\text{m}^2.\)
04

Calculate the Angular Momentum

The angular momentum \(L\) is given by the formula: \(L = I\omega.\) Plug in the values for \(I\) and \(\omega\): \(L = 0.00016875 \times \frac{10\pi}{9}.\) \(L \approx 0.00016875 \times 3.49 \approx 0.0005896\,\text{kg}\cdot\text{m}^2/\text{s}.\)

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Key Concepts

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

Moment of Inertia
When we talk about the moment of inertia, it refers to how difficult it is to change the rotational motion of an object. It’s similar to mass in linear motion, which tells us how hard it is to change the speed of an object. 
For a solid disk, like the record in our problem, the moment of inertia can be calculated using a specific formula: 
  • Formula: \( I = \frac{1}{2}mr^2 \)
  • \( m \) is the mass of the disk
  • \( r \) is the radius
The values given are \( m = 0.015 \, \text{kg} \) and \( r = 0.15 \, \text{m} \). Substituting these into the formula gives:
\( I = \frac{1}{2} \times 0.015 \times (0.15)^2 \)
After computing, we find \( I = 0.00016875 \, \text{kg} \cdot \text{m}^2 \). This measure helps in determining how much torque is needed to get the record rotating at a particular speed.
Angular Speed Conversion
Angular speed tells us how fast something is rotating. In our exercise, it is initially given in revolutions per minute (rpm), which is a common unit in everyday settings like music records. However, for calculations, we prefer radians per second (rad/s).
To convert \( 33 \frac{1}{3} \, \text{rpm} \) into \( \text{rad/s} \), we first handle the fraction:
\( 33 \frac{1}{3} \rightarrow \frac{100}{3} \, \text{rpm} \).
A single revolution is \( 2\pi \) radians, and one minute is \( 60 \) seconds. Therefore, to convert rpm to rad/s we use the conversion factor: \( 1 \, \text{rpm} = \frac{2\pi}{60} \, \text{rad/s} \).
Using this, we find:
\( \omega = \frac{100}{3} \times \frac{2\pi}{60} = \frac{10\pi}{9} \, \text{rad/s} \).
Understanding this conversion is essential for relating real-world rotation speeds to the mathematical models we use in physics.
Solid Disk Model
The solid disk model is a simplified way to understand how objects like records rotate. In this scenario, the disk's mass is distributed evenly across its surface, and we assume the thickness is negligible.
It allows us to use the moment of inertia formula mentioned earlier, helping to calculate rotational properties like angular momentum more easily.
Here’s why this model is useful:
  • Simplicity - It’s easier to calculate the moment of inertia with a simple geometric shape.
  • Accuracy - For many practical purposes, this model gives accurate results.
  • Ease of Understanding - It simplifies complex rotational motion into manageable physics problems.
In our exercise, using the solid disk model made it straightforward to find the angular momentum, which is key for understanding how objects behave when they spin. It highlights the interconnectedness of physics concepts, where understanding one can help unlock the understanding of others.

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Most popular questions from this chapter

A 0.34-kg meterstick balances at its center. If a necklace is suspended from one end of the stick, the balance point moves \(9.5 \mathrm{cm}\) toward that end. (a) Is the mass of the necklace more than, less than, or the same as that of the meterstick? Explain. (b) Find the mass of the necklace.

A baseball bat balances \(71.1 \mathrm{cm}\) from one end. If a \(0.560-\mathrm{kg}\) glove is attached to that end, the balance point moves \(24.7 \mathrm{cm}\) toward the glove. Find the mass of the bat.

Suppose partial melting of the polar ice caps increases the moment of inertia of the Earth from \(0.331 M_{\mathrm{E}} R_{\mathrm{E}}{ }^{2}\) to \(0.332 M_{\mathrm{E}} R_{\mathrm{E}}^{2}\) (a) Would the length of a day (the time required for the Earth to complete one revolution about its axis) increase or decrease? Explain. (b) Calculate the change in the length of a day. Give your answer in seconds.

A turntable with a moment of inertia of \(5.4 \times 10^{-3} \mathrm{kg} \cdot \mathrm{m}^{2}\) rotates freely with an angular speed of \(33 \frac{1}{3}\) rpm. Riding on the rim of the turntable, \(15 \mathrm{cm}\) from the center, is a cute, \(32-g\) mouse. (a) If the mouse walks to the center of the turntable, will the turntable rotate faster, slower, or at the same rate? Explain. (b) Calculate the angular speed of the turntable when the mouse reaches the center.

A person exerts a tangential force of \(36.1 \mathrm{N}\) on the rim of a disk- shaped merry-go-round of radius \(2.74 \mathrm{m}\) and mass \(167 \mathrm{kg}\). If the merry-go-round starts at rest, what is its angular speed after the person has rotated it through an angle of \(32.5^{\circ} ?\)

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