DOC PREVIEW
UT Arlington PHYS 1443 - Lecture Notes

This preview shows page 1-2-3-4 out of 12 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 12 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 12 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 12 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 12 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 12 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

Wednesday, Oct. 2, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu1PHYS 1443 – Section 003Lecture #6Wednesday, Oct. 2, 2002Dr. Jaehoon Yu1. Newton’s laws and its use in uniform and non-uniform circular motion2. Motion in Accelerated Frames3. Motion in Resistive Forces4. Numerical Modeling in Particle Dynamics (Euler Method)Today’s homework is homework #7, due 12:30pm, next Wednesday!!Wednesday, Oct. 2, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu2Announcements• Due time for homework will be changed from 1am to 12pmon the due day, if everyone prefers this…• Term Exam– Exam grading not complete yet. Will be done by next Monday– All scores are relative based on the curve• To take into account the varying difficulties of exams• This average will not be skewed by one or two outliers– Only two best of the three will be used for your final grading, after adjusting each exam scores to the overall average– Exam constitutes only 50% of the total• Do your homework well• Come to the class and do well with quizzesWednesday, Oct. 2, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu3Newton’s Second Law & Uniform Circular MotionFrFrrmrvar2=The centripetal acceleration is always perpendicular to velocity vector, v, for uniform circular motion.The force that causes the centripetal acceleration acts toward the center of the circular path and causes a change in the direction of the velocity vector. This force is called centripetal force.Are there forces in this motion? If so, what do they do?rvmamFrr2==∑What do you think will happen to the ball if the string that holds the ball breaks? Why?Based on Newton’s 1stlaw, since the external force no longer exist, the ball will continue its motion without change and will fly away following the tangential direction to the circle.vWednesday, Oct. 2, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu4Example 6.2A ball of mass 0.500kg is attached to the end of a 1.50m long cord. The ball is moving in a horizontal circle. If the string can withstand maximum tension of 50.0 N, what is the maximum speed the ball can attain before the cord breaks? mFrrvar2=Centripetal acceleration:TrvmamFrr>==∑2When does the string break?When the centripetal force is greater than the sustainable tension.Trvm =2Calculate the tension of the cord when speed of the ball is 5.00m/s.()( )NrvmT 33.85.100.5500.022=×==Ex06-02.ip( )smmTrv /2.12500.05.10.50=×==Wednesday, Oct. 2, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu5Forces in Non-uniform Circular MotionThe object has both tangential and radial accelerations.What does this statement mean?The object is moving under both tangential and radial forces.FrFtFtrFFF +=These forces cause not only the velocity but also the speed of the ball to change. The object undergoes a curved motion under the absence of constraints, such as a string.Wednesday, Oct. 2, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu6Example 6.8A ball of mass m is attached to the end of a cord of length R on Earth. The ball is moving in a vertical circle. Determine the tension of the cord at any instant when the speed of the ball is v and the cord makes an angle θ with vertical. What are the forces involved in this motion?ttmamgF ==∑θsinThe gravitational force Fg and the radial force, T, providing tension. mθRFg=mgAt what angles the tension becomes maximum and minimum. What are the tension?θsingat=RvmmamgTFrr2cos ==−=∑θ+= θcos2gRvmTtangential comp.Radial comp.TθWednesday, Oct. 2, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu7Motion in Accelerated FramesNewton’s laws are valid only when observations are made in an inertial frame of reference. What happens in a non-inertial frame?Fictitious forces are needed to apply Newton’s second law in an accelerated frame.This force does not exist when the observations are made in an inertial reference frame.What does this mean and why is this true?Let’s consider a free ball inside a box under uniform circular motion.We see that the box has a radial force exerted on it but none on the ball directly, until…How does this motion look like in an inertial frame (or frame outside a box)?rFrHow does this motion look like in the box?The ball is tumbled over to the wall of the box and feels that it is getting force that pushes it toward the wall.Why?According to Newton’s first law, the ball wants to continue on its original movement tangentially but since the box is turning, the ball feels like it is being pushed toward the wall relative to everything else in the box.vWednesday, Oct. 2, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu8Non-Inertial FrameExample 6.9A ball of mass m is hung by a cord to the ceiling of a boxcar that is moving with an acceleration a. What do the inertial observer at rest and the non-inertial observer traveling inside the car conclude? How do they differ? mThis is how the ball looks like no matter which frame you are in.TFF g +=∑Inertial FrameθHow do the free-body diagrams look for two frames?Fg=mgmθTFg=mgmθTFficaHow do the motions interpreted in these two frames? Any differences?For an inertial frame observer, the forces being exerted on the ball are only T and Fg. The acceleration of the ball is the same as that of the box car and is provided by the x component of the tension force.ficgFTFF ++=∑In the non-inertial frame observer, the forces being exerted on the ball are T, Fg, and Ffic. For some reason the ball is under a force, Ffic, that provides acceleration to the ball.θsinTmamaFcxx===∑0cos =−=∑mgTFyθθcosmgT =θtangac=0sin =−=∑ficxFTF θ0cos =−=∑mgTFyθθcosmgT =θsinTmaFficfic==θtangafic=While the mathematical expression of the acceleration of the ball is identical to that of inertial frame observer’s, the cause of the force, or physical law is dramatically different.Wednesday, Oct. 2, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu9Motion in Resistive ForcesMedium can exert resistive forces on an object moving through it due to viscosity or other types frictional property of the medium.These forces are exerted on moving objects in opposite direction of the movement. Some examples? These forces are proportional to such factors as speed. They almost always increase with increasing speed. Two different cases of proportionality: 1. Forces linearly proportional to speed: Slowly moving or very small objects2. Forces proportional to square of speed: Large objects w/ reasonable speedAir resistance, viscous force


View Full Document

UT Arlington PHYS 1443 - Lecture Notes

Documents in this Course
Physics

Physics

30 pages

Physics

Physics

25 pages

Physics

Physics

25 pages

Waves

Waves

17 pages

Physics

Physics

16 pages

Friction

Friction

15 pages

Load more
Download Lecture Notes
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Lecture Notes and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Lecture Notes 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?