/*! 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 9 The disk, which has a mass of \(... [FREE SOLUTION] | 91Ó°ÊÓ

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The disk, which has a mass of \(20 \mathrm{kg}\), is subjected to the couple moment of \(M=(2 \theta+4) \mathrm{N} \cdot \mathrm{m},\) where \(\theta\) is in radians. If it starts from rest, determine its angular velocity when it has made two revolutions

Short Answer

Expert verified
The angular velocity after two revolutions is \(\sqrt{16\pi (2\pi + 1)} rad/s\)

Step by step solution

01

Define the Equation of Motion

For rotational motion under a torque, we have the equation of motion as \(I \alpha = M\), where \(I\) is the moment of inertia, \(\alpha\) is the angular acceleration and \(M\) is the torque. For a disk, \(I=0.5*m*r^2\), where \(m\) is the mass and \(r\) the radius. We don't have the radius in this particular problem, but we do know that \(M = 2\theta + 4\). We also have to assume a radius that does not affect the result since it is not given. So, the equation of motion becomes \(0.5*20 \alpha = 2\theta + 4\).
02

Integrate to get Angular Velocity

The equation is not given directly in terms of angular velocity \(\omega\). However, knowing that angular acceleration \(\alpha = d\omega/dt\) and \(\omega = d\theta / dt\), we can rewrite the equation in terms of \(\omega\) as \( d\omega = (2* \theta + 4) dt / 20\). We can integrate this equation from initial angular velocity \(\omega_0\) to final \(\omega_f\) and from initial time \(t_0 = 0\) to final time \(t_f\) to get , \[\int_{\omega_0}^{\omega_f}{\omega d\omega} = \int_{0}^{4\pi}{(2\theta + 4) dt / 20}\]
03

Compute the Required Angular Velocity

Performing the integral, and assuming that the disc starts from rest so \(\omega_0 = 0\), we find that \(\omega_f^2/2 = [(\theta^2 + 4 \theta) 2 / 20]_{0}^{4 \pi} = 8\pi^2 + 16\pi\). Solving for \(\omega_f\) we get \(\omega_f = \sqrt{16\pi (2\pi + 1)} rad/s\).

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Key Concepts

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

Angular Velocity
Angular velocity is a measure of how fast an object rotates around a particular axis. It is the rate of change of angular displacement and is often represented by the Greek letter \( \omega \). If you imagine a wheel spinning, angular velocity describes how quickly it completes each circle. It is measured in radians per second (rad/s), which might sound strange but remember, one full circle creates an angle of \(2\pi\) radians.

When an object starts from rest, like in our exercise, its initial angular velocity \( \omega_0 \) is zero. Over time, as it is subjected to various forces, its angular velocity increases. It is helpful to remember:
  • \( \omega = \frac{d\theta}{dt} \) – Angular velocity is the derivative of angular displacement \( \theta \) with respect to time \( t \).
  • The faster the rotation speed, the higher the angular velocity.
  • If angular velocity increases, angular acceleration is positive; if it decreases, angular acceleration is negative.
Torque and Moment of Inertia
Torque is like a rotational force. It is what causes objects to start rotating, similar to how force causes linear motion. Imagine trying to push a door open. Torque depends on two main things: how much force you apply and how far from the pivot you apply it. The equation for torque \( M \) is given by the formula: \[ M = F \times r \sin(\theta) \]where:
  • \( F \) is the force applied,
  • \( r \) is the distance from the pivot point to where the force is applied,
  • \( \theta \) is the angle between the force and the arm.


The moment of inertia \( I \) resists these changes in motion, much like mass resists linear acceleration. It's a property that depends on the object's mass and shape. For a disk, the moment of inertia is usually calculated as \( I = 0.5 \times m \times r^2 \). This property determines how much torque is needed to induce a given angular acceleration \( \alpha \). The relationship between torque, moment of inertia, and angular acceleration is quite simple:
  • \( I \alpha = M \)
  • More moment of inertia means more torque is needed to rotate the object.
Rotational Motion Equations
Just like there are equations for linear motion, rotational motion has its set of equations. These equations help describe how an object moves through an angle over time. In our problem, understanding rotational motion is key. If you know the torque acting on an object and its moment of inertia, you can find out how its rotational speed changes.

There are several important equations:
  • \( \alpha = \frac{d\omega}{dt} \) – Angular acceleration is the rate of change of angular velocity.
  • \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \) – This is the rotational analog of the linear motion equation \( s = ut + \frac{1}{2}at^2 \).
  • Torque plays a crucial role as it links angular acceleration and moment of inertia: \( I \alpha = M \).
  • Combining these, \( \omega \) changes are computed as in: \( \omega^2 = \omega_0^2 + 2\alpha \theta \)
In our original problem, computing the angular velocity involves integrating the angular acceleration over the path of motion.

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Most popular questions from this chapter

If the gear is released from rest, determine its angular velocity after its center of gravity \(O\) has descended a distance of \(4 \mathrm{ft}\). The gear has a weight of \(100 \mathrm{lb}\) and a radius of gyration about its center of gravity of \(k=0.75 \mathrm{ft}\)

The system consists of a \(30-\mathrm{kg}\) disk \(A, 12-\mathrm{kg}\) slender rod \(B A\), and a \(5-\mathrm{kg}\) smooth collar \(A\). If the disk rolls without slipping, determine the velocity of the collar at the instant \(\theta=30^{\circ} .\) The system is released from rest when \(\theta=45^{\circ}\)

The 10 -lb sphere starts from rest at \(\theta=0^{\circ}\) and rolls without slipping down the cylindrical surface which has a radius of \(10 \mathrm{ft}\). Determine the speed of the sphere's center of mass at the instant \(\theta=45^{\circ}\)

A force of \(P=20 \mathrm{N}\) is applied to the cable, which causes the \(175-\mathrm{kg}\) reel to turn without slipping on the two rollers \(A\) and \(B\) of the dispenser. Determine the angular velocity of the reel after it has made two revolutions starting from rest. Neglect the mass of the cable. Each roller can be considered as an 18 -kg cylinder, having a radius of \(0.1 \mathrm{m} .\) The radius of gyration of the reel about its center axis is \(k_{G}=0.42 \mathrm{m}\)

. The wheel has a mass of \(100 \mathrm{kg}\) and a radius of gyration of \(k_{O}=0.2 \mathrm{m} .\) A motor supplies a torque \(M=(40 \theta+900) \mathrm{N} \cdot \mathrm{m},\) where \(\theta\) is in radians, about the drive shaft at \(O .\) Determine the speed of the loading car, which has a mass of \(300 \mathrm{kg}\), after it travels \(s=4 \mathrm{m}\). Initially the car is at rest when \(s=0\) and \(\theta=0^{\circ} .\) Neglect the mass of the attached cable and the mass of the car's wheels.

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