/*! 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 61 The semicircular disk of mass \(... [FREE SOLUTION] | 91Ó°ÊÓ

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The semicircular disk of mass \(m\) and radius \(r\) is released from rest at \(\theta=0\) and rotates freely in the vertical plane about its fixed bearing at \(O .\) Derive expressions for the \(n\) - and \(t\) -components of the force \(F\) on the bearing as functions of \(\theta\)

Short Answer

Expert verified
The force component expressions are \(F_n = \frac{4mgr}{3\pi}\omega^2\) and \(F_t = mg\cos\theta\).

Step by step solution

01

Identify the forces and motion

Consider the semicircular disk that is free to rotate about a fixed point at the center of the flat edge labeled as point \(O\). The gravitational force acts through the center of mass of the semicircle. We'll denote the forces acting on the bearing as \(F_n\) (normal) and \(F_t\) (tangential) components.
02

Determine location of center of mass

The center of mass of a semicircular disk with radius \(r\) is located at a distance \(\frac{4r}{3\pi}\) from the center of the circle along the axis of symmetry. We'll use this in calculating the torques and forces.
03

Calculate velocity of center of mass

The angular velocity \(\omega\) as a function of \(\theta\) can be obtained using energy conservation: initial potential energy converts to rotational kinetic energy. \[ mgh = \frac{1}{2}I\omega^2 \] where \(h = \frac{4r}{3\pi}(1 - \cos\theta)\), and \(I = \frac{1}{2}mr^2\) for a semicircle. Solve for \(\omega\).
04

Apply dynamics for rotating body

Using the component of gravity that causes rotation, \(mg\sin\theta\), relate this to \(I\alpha\) (torque equals moment of inertia times angular acceleration) to find \(\alpha\). For a rotating system without slipping, \(\alpha\) and \(\omega\) are related through angular displacement \(\theta\).
05

Solve for normal and tangential forces

Using the circular motion principles, \(F_n = m\cdot \omega^2\cdot\frac{4r}{3\pi}\) provides the radial (centripetal) force, resulting from gravitational and centripetal influences. \(F_t = mg\cos\theta - 0\) (since no external tangential forces except gravity act in the \(t\)-direction) can be calculated using net forces in the tangential direction.
06

Final expressions for \(n\) and \(t\) components

Combine these equations: \(F_n = \frac{4mgr}{3\pi}\left( \omega^2 \right)\) and \(F_t = mg\cos\theta\). Express \(\omega\) in terms of \(\theta\) using values from Step 3 to complete the expressions.

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Key Concepts

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

Dynamics
Dynamics is the study of forces and motion. In this exercise, we're looking at how a semicircular disk moves and the forces that act upon it as it rotates. When the disk is released, it begins rotating around a fixed point due to gravitational forces. To analyze its motion, we break down the forces into components. These components are the normal component, acting perpendicular to the motion, and the tangential component, acting along the motion's path.
  • The normal component of force ( F_n) is associated with centripetal effects and constraints of the motion path.
  • The tangential component of force ( F_t) is related to the acceleration along the path tangent to the circular motion.
Dynamics helps us quantify these forces by relating them to motion through equations such as Newton's Second Law. In this exercise, we are particularly interested in how gravitational force contributes to these components, enhancing our understanding of rotational motion and energy conservation.
Center of Mass
The center of mass is a crucial concept when analyzing any object's motion. For a semicircular disk, the center of mass is not located at its geometric center, as it would be for a full circle. Instead, it is positioned along the axis of symmetry at a distance \(\frac{4r}{3\pi}\) from the center of the circle. This position affects how the disk behaves under rotational motion.
Knowing the location of the center of mass allows us to calculate torques and forces as the disk rotates. In practice, the center of mass acts as the effective point where gravitational forces apply, influencing both the rotational kinetic energy and the dynamics of the system. It's a vital point of reference for simplifying complex motion problems and finding accurate solutions.
Rotational Motion
Rotational motion describes how an object moves when it spins around a fixed axis. In this scenario, a semicircular disk rotates about the point O, which is fixed. As the disk falls, it gains angular velocity \(\omega\).
  • Angular velocity describes how fast the disk spins per unit time. It's connected to rotational kinetic energy as it indicates the energy due to rotation.
  • Angular acceleration \(\alpha\) is the rate of change of angular velocity. It reflects how quickly the object picks up speed when rotating.
For a disk, its moment of inertia \(I\) plays a role in these calculations. It's computed as \(I = \frac{1}{2}mr^2\). This factor influences how the disk accelerates under applied torques, especially the gravitational torque \(mg\sin\theta\), which induces its rotational motion.
Energy Conservation
Energy conservation is a key principle that applies to this problem. At the start, the semicircular disk has potential energy due to its height in the gravitational field. As it begins to rotate, that potential energy converts into kinetic energy, which can be either linear or rotational.
  • Initially, the potential energy is given by \(mgh\), where \(h = \frac{4r}{3\pi}(1 - \cos\theta)\). This equation considers the drop in height as the disk swings down.
  • Rotational kinetic energy is computed by \(\frac{1}{2}I\omega^2\). This formula describes the energy the object possesses due to its rotation.
The principles of energy conservation allow us to relate the initial and final states of the system without any external work being done. The conversion from potential to kinetic energy ensures a smooth understanding of how \(\omega\) depends on the angle \(\theta\). It's a beautiful illustration of energy principles directly applied to rotating systems.

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Most popular questions from this chapter

The sector and attached wheels are released from rest in the position shown in the vertical plane. Each wheel is a solid circular disk weighing \(12 \mathrm{lb}\) and rolls on the fixed circular path without slipping. The sector weighs 18 lb and is closely approximated by one-fourth of a solid circular disk of 16-in. radius. Determine the initial angular acceleration \(\alpha\) of the sector.

A person who walks through the revolving door exerts a 90 -N horizontal force on one of the four door panels and keeps the \(15^{\circ}\) angle constant relative to a line which is normal to the panel. If each panel is modeled by a 60 -kg uniform rectangular plate which is \(1.2 \mathrm{m}\) in length as viewed from above, determine the final angular velocity \(\omega\) of the door if the person exerts the force for 3 seconds. The door is initially at rest and friction may be neglected.

The uniform 225 -lb crate is supported by the thin homogeneous 40 -lb platform \(B F\) and light support links whose motion is controlled by the hydraulic cylinder \(C D .\) If the cylinder is extending at a constant rate of 6 in./sec when \(\theta=75^{\circ},\) determine the magnitudes of the forces supported by the pins at \(B\) and \(F\). Additionally, determine the total friction force acting on the crate. The crate is centered on the 6 -ft platform, and friction is sufficient to keep the crate motionless relative to the platform. (Hint: Be careful with the location of the resultant normal force beneath the crate.)

The 50 -kg flywheel has a radius of gyration \(\bar{k}=0.4 \mathrm{m}\) about its shaft axis and is subjected to the torque \(M=2\left(1-e^{-0.1 \theta}\right) \mathrm{N} \cdot \mathrm{m},\) where \(\theta\) is in radians. If the flywheel is at rest when \(\theta=0,\) determine its angular velocity after 5 revolutions.

Under active development is the storage of energy in high-speed rotating disks where friction is effectively eliminated by encasing the rotor in an evacuated enclosure and by using magnetic bearings. For a 10 -kg rotor with a radius of gyration of \(90 \mathrm{mm}\) rotating initially at \(80000 \mathrm{rev} / \mathrm{min},\) calculate the power \(P\) which can be extracted from the rotor by applying a constant \(2.10-\mathrm{N} \cdot \mathrm{m}\) retarding torque \((a)\) when the torque is first applied and ( \(b\) ) at the instant when the torque has been applied for 120 seconds.

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