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Chapter 11: Problem 38

A sanding disk with rotational inertia \(1.2 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\) is attached to an electric drill whose motor delivers a torque of magnitude \(16 \mathrm{~N} \cdot \mathrm{m}\) about the central axis of the disk. About that axis and with the torque applied for \(33 \mathrm{~ms}\), what is the magnitude of the (a) angular momentum and (b) angular velocity of the disk?

Short Answer

Expert verified
(a) The angular momentum is 0.528 kg路m虏/s. (b) The angular velocity is 440 rad/s.

Step by step solution

01

Understand the Problem

The problem involves calculating the angular momentum and angular velocity of a rotating object. The given rotational inertia is \(1.2 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\), torque is \(16 \mathrm{~N} \cdot \mathrm{m}\), and time of torque application is \(33 \mathrm{~ms}\).
02

Identify Formulas

To find angular momentum (\(L\)), we use the formula \(L = I \cdot \omega\), where \(I\) is rotational inertia, and \(\omega\) is angular velocity. To find angular velocity, we use \(\omega = \frac{\tau \cdot t}{I}\), where \(\tau\) is torque and \(t\) is time.
03

Convert Units

Convert the time from milliseconds to seconds for consistency with SI units: \(33 \mathrm{~ms} = 0.033 \mathrm{~s}\).
04

Calculate Angular Momentum

First, calculate angular velocity \(\omega\) using the formula \(\omega = \frac{\tau \cdot t}{I}\). Substitute the given values: \(\omega = \frac{16 \cdot 0.033}{1.2 \times 10^{-3}}\). Then calculate \(L = I \cdot \omega\) using the calculated \(\omega\).
05

Solve for Angular Velocity

Compute \(\omega = \frac{16 \cdot 0.033}{1.2 \times 10^{-3}} = \frac{0.528}{1.2 \times 10^{-3}} = 440 \mathrm{~rad/s}\).
06

Solve for Angular Momentum

Using calculated \(\omega\), find \(L = 1.2 \times 10^{-3} \cdot 440 = 0.528 \mathrm{~kg} \cdot \mathrm{m}^{2}/s\).

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Key Concepts

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

Angular Momentum
Angular momentum is a measure of the quantity of rotation of an object and is an important concept in physics. It is conserved in a system, unless acted upon by an external torque. The angular momentum (\(L\)) of a rotating body is calculated by multiplying the rotational inertia (\(I\)) by the angular velocity (\(\omega\)):
  1. Formula: \[L = I \times \omega\]
  2. Units: It is measured in kilogram meter squared per second (\(\mathrm{kg} \cdot \mathrm{m}^2/\mathrm{s}\)).
For example, in our exercise, once we calculated the angular velocity, we used it to determine the angular momentum using the given rotational inertia of the disk. By understanding this fundamental relationship, you can solve similar rotational dynamics problems easily.
Angular Velocity
Angular velocity represents how fast an object rotates or revolves relative to another point鈥攕pecifically, how quickly the angular position or orientation of an object changes with time. Unlike linear velocity, which refers to motion along a straight path, angular velocity refers to rotational motion around an axis.
  • Formula: The angular velocity can be determined by the formula \[\omega = \frac{\tau \cdot t}{I}\], where \(\tau\) is the torque applied, \(t\) is the time duration, and \(I\) is the rotational inertia.
  • Units: Measured in radians per second (\(\mathrm{rad/s}\)).
In the original exercise, the torque applied by the drill was used along with the time to calculate the angular velocity of the disk. This is a key step in understanding how rotational dynamics combine these elements to describe motion.
Rotational Inertia
Rotational inertia, also known as the moment of inertia, is a property of any object that defines how difficult it is to change its rotational motion. It depends on both the mass of the object and the distribution of mass around the axis of rotation.
  • Formula: It is used directly in calculations of angular momentum and angular velocity as \(I\) in both formulas.
  • Units: The units for rotational inertia are kilogram meter squared (\(\mathrm{kg} \cdot \mathrm{m}^2\)).
In our exercise, the rotational inertia value was provided, allowing us to solve for the angular velocity and, subsequently, angular momentum. This concept is especially important in rotational dynamics, as it shows how an object's mass distribution affects its resistance to changes in motion.

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Most popular questions from this chapter

A uniform block of granite in the shape of a book has face dimensions of \(20 \mathrm{~cm}\) and \(15 \mathrm{~cm}\) and a thickness of \(1.2 \mathrm{~cm} .\) The density (mass per unit volume) of granite is \(2.64 \mathrm{~g} / \mathrm{cm}^{3} .\) The block rotates around an axis that is perpendicular to its face and halfway between its center and a corner. Its angular momentum about that axis is \(0.104 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\). What is its rotational kinetic energy about that axis?

A disk with a rotational inertia of \(7.00 \mathrm{~kg} \cdot \mathrm{m}^{2}\) rotates like a merry-go-round while undergoing a time-dependent torque given by \(\tau=(5.00+2.00 t) \mathrm{N} \cdot \mathrm{m} .\) At time \(t=1.00 \mathrm{~s},\) its angular momentum is \(5.00 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\). What is its angular momentum at \(t=3.00 \mathrm{~s} ?\)

A body of radius \(R\) and mass \(m\) is rolling smoothly with speed \(v\) on a horizontal surface. It then rolls up a hill to a maximum height \(h\). (a) If \(h=3 v^{2} / 4 g\), what is the body's rotational inertia about the rotational axis through its center of mass? (b) What might the body be?

In unit-vector notation, what is the torque about the origin on a jar of jalapeno peppers located at coordinates \((3.0 \mathrm{~m},-2.0 \mathrm{~m}\), \(4.0 \mathrm{~m})\) due to (a) force \(\vec{F}_{1}=(3.0 \mathrm{~N}) \mathrm{i}-(4.0 \mathrm{~N}) \mathrm{j}+(5.0 \mathrm{~N}) \mathrm{k}\) (b) force \(\vec{F}_{2}=(-3.0 \mathrm{~N}) \hat{\mathrm{i}}-(4.0 \mathrm{~N}) \mathrm{j}-(5.0 \mathrm{~N}) \mathrm{k},\) and (c) the vector sum of \(\vec{F}_{1}\) and \(\vec{F}_{2} ?\) (d) Repeat part (c) for the torque about the point with coordinates \((3.0 \mathrm{~m}, 2.0 \mathrm{~m}, 4.0 \mathrm{~m})\).

In unit-vector notation, what is the net torque about the origin on a flea located at coordinates \((0,-4.0 \mathrm{~m}, 5.0 \mathrm{~m})\) when forces \(\vec{F}_{1}=(3.0 \mathrm{~N}) \mathrm{k}\) and \(\vec{F}_{2}=(-2.0 \mathrm{~N}) \mathrm{j}\) act on the flea?
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