Problem 38
Question
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?
Step-by-Step Solution
Verified Answer
(a) The angular momentum is 0.528 kg·m²/s. (b) The angular velocity is 440 rad/s.
1Step 1: 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 \(33 \mathrm{~ms}\).
2Step 2: 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.
3Step 3: Convert Units
Convert the time from milliseconds to seconds for consistency with SI units: \(33 \mathrm{~ms} = 0.033 \mathrm{~s}\).
4Step 4: 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 \(\omega\).
5Step 5: Solve for Angular Velocity
Compute \(\omega = \frac{16 \cdot 0.033}{1.2 \times 10^{-3}} = \frac{0.528}{1.2 \times 10^{-3}} = 440 \mathrm{~rad/s}\).
6Step 6: Solve for Angular Momentum
Using calculated \(\omega\), find \(L = 1.2 \times 10^{-3} \cdot 440 = 0.528 \mathrm{~kg} \cdot \mathrm{m}^{2}/s\).
Key Concepts
Angular MomentumAngular VelocityRotational Inertia
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 (\(\omega\)):
- Formula: \[L = I \times \omega\]
- Units: It is measured in kilogram meter squared per second (\(\mathrm{kg} \cdot \mathrm{m}^2/\mathrm{s}\)).
Angular Velocity
Angular velocity represents how fast an object rotates or revolves relative to another point—specifically, 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 (\(\mathrm{rad/s}\)).
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 (\(\mathrm{kg} \cdot \mathrm{m}^2\)).
Other exercises in this chapter
Problem 33
At time \(t=0,\) a \(3.0 \mathrm{~kg}\) particle with velocity \(\vec{v}=(5.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}-(6.0 \mathrm{~m} / \mathrm{s}) \hat{\ma
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Three Particles of mass \(m=23 \mathrm{~g}\) are fastened to three rods of length \(d=12 \mathrm{~cm}\) and negligible mass. The rigid assembly rotates around p
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The angular momentum of a flywheel having a rotational inertia of \(0.140 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about its central axis decreases from 3.00 to \(0.
View solution Problem 40
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 \
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