/*! 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 126 The 20 -kg wheel has an eccentri... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The 20 -kg wheel has an eccentric mass which places the center of mass \(G\) a distance \(\bar{r}=70 \mathrm{mm}\) away from the geometric center \(0 .\) A constant couple \(M=6 \mathrm{N} \cdot \mathrm{m}\) is applied to the initially stationary wheel, which rolls without slipping along the horizontal surface and enters the curve of radius \(R=600 \mathrm{mm} .\) Determine the normal force under the wheel just before it exits the curve at \(C .\) The wheel has a rolling radius \(r=100 \mathrm{mm}\) and a radius of gyration \(k_{O}=65 \mathrm{mm}\)

Short Answer

Expert verified
Calculate forces on the wheel, determine accelerations and solve for the normal force using motion equations.

Step by step solution

01

Analyze the Forces on the Wheel

The wheel experiences gravity, the normal force, friction, and the applied couple. Begin by identifying that the normal force is vertical, acting upwards through the contact point with the ground.
02

Determine Moment of Inertia

Calculate the wheel's moment of inertia about the center using the radius of gyration:\[ I_O = m imes k_O^2 = 20 imes (0.065)^2 = 0.0845 \, \text{kg} \cdot \text{m}^2 \]
03

Apply Newton's Second Law for Rotation

Using Newton's Second Law for rotation, write the equation for the angular acceleration \( \alpha \):\[ M = I_O \times \alpha \rightarrow 6 = 0.0845 \times \alpha \]This gives us \( \alpha = \frac{6}{0.0845} = 71.006 \text{ rad/s}^2 \).
04

Determine Linear Acceleration of the Center of Mass

Relate angular acceleration to linear acceleration:\[ a_G = \alpha \times r = 71.006 \times 0.1 = 7.1006 \text{ m/s}^2 \]
05

Calculate the Centripetal Acceleration

At point \(C\), the centripetal acceleration \(a_c\) is given by \(a_c = \frac{v^2}{R}\). Use the geometric relationship that velocity \(v = \omega \times r\) and that \( \alpha \) is the rate of change of \( \omega \), where \( \omega = \alpha \times t \) since the wheel is initially at rest.
06

Determine Angular Velocity at Point C

Using angular acceleration and time of rolling till the point just before exit:If \( t \) is the time taken to reach point C:\[ v = r \alpha t \rightarrow \omega = \alpha t \]This value is used to find \(a_c\).
07

Calculate the Normal Force

Use the equation of motion in the vertical direction. The normal force \( N \) is given by:\[ N = m g - m rac{v^2}{R} + m a_G \]Substitute known values and solve for \( N \):\[ N = 20 \times 9.81 - 20 \times \frac{(r\omega)^2}{600} + 20 \times 7.1006 \]
08

Simplify and Solve

Calculate speeds and substitute to find the solution:Substitute \(v=r\omega\) and solve for normal force \(N\) incorporating approximate numerical values.

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.

Dynamics
Dynamics is a branch of physics that deals with the motion of objects and the forces that affect this motion. In the context of this exercise, we are dealing with rotational dynamics. This is particularly interesting as we analyze not just the translational movement of the wheel, but also its rotational behavior caused by the applied couple. When a couple is applied, it induces angular acceleration, changing the wheel's rotational speed.
Each component of the system is essential to determine the final result, especially the forces involved. These include gravitational force acting downward, the normal reaction force from the surface, and friction that is necessary to prevent slipping. All these forces must be considered in calculating how the wheel moves along the surface and through the curve.
Understanding these concepts requires applying Newton's laws of motion. Specifically, Newton's Second Law is instrumental in relating forces to the linear and angular accelerations experienced by the body.
Centripetal Force
Centripetal force is the inward force required to keep an object moving in a circular path. When the wheel enters and travels through the curve, it requires a centripetal force to maintain its circular motion. This force is directed towards the center of the curve.
The expression for centripetal force \[ F_c = rac{mv^2}{R} \] must be understood as a result of the requirement to change direction. The faster an object moves or the tighter the curve, the greater the force needed to stay in the circular path. For the wheel, this centripetal force is partly provided by the friction and partly by the normal force.
When the problem states we need to find the normal force just as the wheel exits the curve, it highlights the dynamic importance of centripetal force transitioning back to purely normal, as straight motion requires none.
Angular Acceleration
Angular acceleration is a measure of how quickly a rotating object’s speed changes. In this problem, angular acceleration is particularly crucial because a constant couple is applied to the wheel. This couple leads to a torque that causes the wheel to accelerate from rest.
The formula for angular acceleration is \[ \alpha = \frac{M}{I_O} \] where \( M \) is the applied torque, and \( I_O \) is the moment of inertia. From the solution, we know this angular acceleration is used to figure out the wheel's rotational velocity as it moves along the curved path.
This acceleration directly affects linear acceleration as well, as they are related by the equation \( a_G = \alpha \times r \), ensuring that the speed of rotation corresponds to the forward movement of the wheel. This interplay between angular and linear motion is foundational in understanding rotational dynamics.
Normal Force
Normal force is the perpendicular force exerted by a surface to support the weight of an object resting on it, here it's the wheel on the ground. Calculating the normal force acting on the wheel as it exits the curve involves combining several aspects of dynamics and motion.
The total normal force is influenced by the wheel's weight, leniently supported minus what is required as centripetal force, coupled with the horizontal force due to rolling induced acceleration of the wheel’s center. Thus, the equation used for the problem is: \[ N = mg - m\frac{v^2}{R} + ma_G \] indicating that normal force not only upholds the weight but also adjusts according to the wheel’s acceleration and velocity.
Through analyzing these forces, students can appreciate the delicate balance governing motion over curbed tracks. The method of breaking down the sum of forces in each direction fosters a comprehensive understanding of how dynamics shape real-world mechanical systems.

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

The motor \(M\) is used to hoist the 12,000 -lb stadium panel (centroidal radius of gyration \(\bar{k}=6.5 \mathrm{ft}\) ) into position by pivoting the panel about its corner \(A\). If the motor is capable of producing 5000 lb-ft of torque, what pulley diameter \(d\) will give the panel an initial counterclockwise angular acceleration of \(1.5 \mathrm{deg} / \mathrm{sec}^{2} ?\) Neglect all friction.

The 75 -kg flywheel has a radius of gyration about its shaft axis of \(\bar{k}=0.50 \mathrm{m}\) and is subjected to the torque \(M=10\left(1-e^{-t}\right) \mathrm{N} \cdot \mathrm{m},\) where \(t\) is in seconds. If the flywheel is at rest at time \(t=0\) determine its angular velocity \(\omega\) at \(t=3\) s.

A flexible cable 60 meters long with a mass of \(0.160 \mathrm{kg}\) per meter of length is wound around the reel. With \(y=0,\) the weight of the 4 -kg cylinder is required to start turning the reel to overcome friction in its bearings. Determine the downward acceleration \(a\) in meters per second squared of the cylinder as a function of \(y\) in meters. The empty reel has a mass of \(16 \mathrm{kg}\) with a radius of gyration about its bearing of \(200 \mathrm{mm}\)

The uniform slender bar \(A B\) has a mass of \(8 \mathrm{kg}\) and swings in a vertical plane about the pivot at \(A\). If \(\dot{\theta}=2 \mathrm{rad} / \mathrm{s}\) when \(\theta=30^{\circ},\) compute the force supported by the pin at \(A\) at that instant.

The 64.4 -lb solid circular disk is initially at rest on the horizontal surface when a 3 -lb force \(P\), constant in magnitude and direction, is applied to the cord wrapped securely around its periphery. Friction between the disk and the surface is negligible. Calculate the angular velocity \(\omega\) of the disk after the \(3-1 b\) force has been applied for 2 seconds and find the linear velocity \(v\) of the center of the disk after it has moved 3 feet from rest.

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.