/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 40 A disk with a rotational inertia... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 \(t=3.00 \mathrm{~s}\) ?

Short Answer

Expert verified
The angular momentum at \( t = 3.00 \) s is \( 23.00 \, \mathrm{kg} \cdot \mathrm{m}^2/\mathrm{s} \).

Step by step solution

01

Understand the Angular Momentum Relationship

The relationship between torque \( \tau \) and angular momentum \( L \) is given by the equation \( \frac{dL}{dt} = \tau \). This means the rate of change of angular momentum is equal to the applied torque.
02

Integrate the Torque over Time

To find the change in angular momentum \( \Delta L \), integrate the torque function \( \tau(t) = 5.00 + 2.00t \) over the time interval from \( t = 1.00 \) s to \( t = 3.00 \) s. \[\Delta L = \int_{1}^{3} (5.00 + 2.00t) \, dt.\]
03

Calculate the Integral

Perform the integration:\[\Delta L = \left[ 5.00t + t^2 \right]_{1}^{3} = \left(5.00(3) + (3)^2\right) - \left(5.00(1) + (1)^2\right).\]Simplify:\[\Delta L = (15 + 9) - (5 + 1) = 24 - 6 = 18 \, \mathrm{kg} \cdot \mathrm{m}^2/\mathrm{s}.\]
04

Calculate the Angular Momentum at t=3.00s

The initial angular momentum at \( t = 1.00 \) s is \( L_1 = 5.00 \, \mathrm{kg} \cdot \mathrm{m}^2/\mathrm{s} \). Add the change in angular momentum:\[L_2 = L_1 + \Delta L = 5.00 + 18 = 23.00 \, \mathrm{kg} \cdot \mathrm{m}^2/\mathrm{s}.\]Thus, the angular momentum at \( t = 3.00 \) s is \( 23.00 \, \mathrm{kg} \cdot \mathrm{m}^2/\mathrm{s} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Rotational Inertia
Rotational inertia, also known as moment of inertia, is a fundamental concept in understanding rotational motion. It's a measure of how difficult it is to change the rotational state of an object. Imagine it as the rotational equivalent of mass in linear motion. For a disk like our merry-go-round example, the rotational inertia depends on both the mass distribution and the axis about which the object rotates.
This concept is crucial when calculating the effect of applied forces, such as torque, because it affects how quickly or slowly an object will start to spin. The greater the rotational inertia, the harder it is to change the angular velocity of the object.
In practical terms, this means that a heavier or more widely distributed mass, like a large disc, will have greater inertia and is more resistant to rotational change. This is why the same amount of torque will cause different angular accelerations depending on the object's rotational inertia.
Torque Integration
Torque, represented by the symbol \( \tau \), is the rotational analog of force in linear dynamics. It is what causes an object to rotate and is dependent on factors such as the force applied and the distance from the pivot point.
In dynamic systems where torque is not constant over time, as in our problem \(\tau = 5.00 + 2.00t \, \mathrm{N} \cdot \mathrm{m}\), integration is used to find the total change in angular momentum over a specified time period. By integrating the torque function over time, we calculate the cumulative effect of torque on the object.
The integral gives the total change in angular momentum from one time to another:
  • This approach involves finding the area under the torque vs. time curve, which represents how much angular momentum has been added or removed.
  • In the problem above, integration over the interval from \( t = 1.00 \) s to \( t = 3.00 \) s, results in an angular momentum change of \( 18 \, \mathrm{kg} \cdot \mathrm{m}^2/\mathrm{s} \).
Angular Dynamics
Angular dynamics refers to the study of rotating objects under the influence of torques. It's the counterpart to linear dynamics but applied to rotation. The key relationships include how torque influences the angular momentum and angular acceleration of an object.
The primary equation governing these dynamics is \( \tau = I \cdot \alpha \), where \( \tau \) is the torque, \( I \) is the rotational inertia, and \( \alpha \) is the angular acceleration. This shows us that torque causes an object to change its rotational state by generating angular acceleration.
When dealing with time-dependent torques, the connection with angular momentum becomes clear as the rate of change of angular momentum \((\frac{dL}{dt})\) is equal to the torque.
This concept helps explain how various torques affect the motion of rotating bodies and is fundamental in understanding complex rotational mechanisms in engineering and physics scenarios.
Time-Dependent Torque
Time-dependent torque can significantly influence the rotation of a body, making problems engaging by considering how torque changes with time. In our example, the torque increases linearly with time \((\tau = 5.00 + 2.00t)\), which means that as time increases, more torque is being applied to the disk.
This type of torque is crucial when analyzing real-world situations where the applied forces and their leverage can change, such as in motor vehicles or machinery that speeds up or slows down dynamically.
By understanding how the torque varies, one can accurately determine the rotational behavior of the object over time. It also helps engineers design mechanical systems that can adapt to varying conditions, providing both safety and efficiency.
  • In mathematical problems, time-dependent torque means that integration is necessary to capture its cumulative effect over time, as seen in the exercise solution.
  • This results in changes to the angular momentum that cannot simply be calculated by multiplying torque by time, stressing the importance of calculus in physics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A bowler throws a bowling ball of radius \(R=11 \mathrm{~cm}\) along a lane. The ball (Fig. \(11-38\) ) slides on the lane with initial speed \(v_{\mathrm{com}, 0}=8.5 \mathrm{~m} / \mathrm{s}\) and initial angular speed \(\omega_{0}=0 .\) The coefficient of kinetic friction between the ball and the lane is \(0.21\). The kinetic frictional force \(\vec{f}_{k}\) acting on the ball causes a linear acceleration of the ball while producing a torque that causes an angular acceleration of the ball. When speed \(v_{\text {com }}\) has decreased enough and angular speed \(\omega\) has increased enough, the ball stops sliding and then rolls smoothly. (a) What then is \(v_{\text {com }}\) in terms of \(\omega\) ? During the sliding, what are the ball's (b) linear acceleration and (c) angular acceleration? (d) How long does the ball slide? (e) How far does the ball slide? (f) What is the linear speed of the ball when smooth rolling begins?

Aparticle moves through an \(x y z\) coordinate system while a force acts on the particle. When the particle has the position vector \(\vec{r}=(2.00 \mathrm{~m}) \hat{\mathrm{i}}-(3.00 \mathrm{~m}) \hat{\mathrm{j}}+(2.00 \mathrm{~m}) \hat{\mathrm{k}}\), the force is given by \(\vec{F}=F_{x} \hat{\mathrm{i}}+(7.00 \mathrm{~N}) \hat{\mathrm{j}}-(6.00 \mathrm{~N}) \hat{\mathrm{k}}\) and the corresponding torque about the origin is \(\vec{\tau}=(4.00 \mathrm{~N} \cdot \mathrm{m}) \hat{\mathrm{i}}+(2.00 \mathrm{~N} \cdot \mathrm{m}) \hat{\mathrm{j}}-(1.00 \mathrm{~N} \cdot \mathrm{m}) \hat{\mathrm{k}}\). Determine \(F_{x^{*}}\)

In a playground, there is a small merry-go-round of radius \(1.20 \mathrm{~m}\) and mass \(180 \mathrm{~kg} .\) Its radius of gyration (see Problem 79 of Chapter 10 ) is \(91.0 \mathrm{~cm}\). A child of mass \(44.0 \mathrm{~kg}\) runs at a speed of \(3.00 \mathrm{~m} / \mathrm{s}\) along a path that is tangent to the rim of the initially stationary merry-go-round and then jumps on. Neglect friction between the bearings and the shaft of the merry-go-round. Calculate (a) the rotational inertia of the merry-go-round about its axis of rotation, (b) the magnitude of the angular momentum of the running child about the axis of rotation of the merry-go-round, and (c) the angular speed of the merry-go- round and child after the child has jumped onto the merry-go-round.

The rotor of an electric motor has rotational inertia \(I_{m}=\) \(2.0 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\) about its central axis. The motor is used to change the orientation of the space probe in which it is mounted. The motor axis is mounted along the central axis of the probe; the probe has rotational inertia \(I_{p}=12 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about this axis. Calculate the number of revolutions of the rotor required to turn the probe through \(30^{\circ}\) about its central axis.

A \(2.50 \mathrm{~kg}\) particle that is moving horizontally over a floor with velocity \((-3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\) undergoes a completely inelastic collision with a \(4.00 \mathrm{~kg}\) particle that is moving horizontally over the floor with velocity \((4.50 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}\). The collision occurs at \(x y\) coordinates \((-0.500 \mathrm{~m},-0.100 \mathrm{~m})\). After the collision and in unit- vector notation, what is the angular momentum of the stuck-together particles with respect to the origin?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.