Your SlideShare is downloading. ×
Kinematics lecture
Upcoming SlideShare
Loading in...5
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

×
Saving this for later? Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime – even offline.
Text the download link to your phone
Standard text messaging rates apply

Kinematics lecture

613

Published on

Published in: Technology, Education
0 Comments
2 Likes
Statistics
Notes
  • Be the first to comment

No Downloads
Views
Total Views
613
On Slideshare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
Downloads
43
Comments
0
Likes
2
Embeds 0
No embeds

Report content
Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
No notes for slide

Transcript

  • 1. Exam IPhysics 101: Lecture 03 Kinematics Today’s lecture will cover Textbook Sections 3.1-3.3 (and some Ch. 4) Physics 101: Lecture 3, Pg 1
  • 2. AnnouncementsHW 1 is due Tuesday Aug 31 (tomorrow) at 6am.No clicker points for Lect. 2? Re-register.Office hours now underway.Read the course description & FAQ on the course web site! Physics 101: Lecture 3, Pg 2
  • 3. Force at Angle Example A person is pushing a 15 kg block across a floor with mk= 0.4 at a constant speed. If she is pushing down at an angle of 25 degrees, what is the magnitude of her force on the block?x- direction: SFx = max Combine: Fpush cos(q) – Ffriction = 0 Fpush cos(q) / m–mg – FPush sin(q) = 0 Fpush cos(q) – m FNormal = 0 Fpush ( cos(q) / m - sin(q)) = mg FNormal = Fpush cos(q) / m Fpush = m g / ( cos(q)/m – sin(q))y- direction: SFy = may Fpush = 80 NFNormal –Fweight – FPush sin(q) = 0 NormalFNormal –mg – FPush sin(q) = 0 Pushing y q x q Friction Weight Physics 101: Lecture 3, Pg 3
  • 4. Homework 2 Example Calculate the tension in the left string. Work through example? y A: Yes B: No x x-direction: SF=ma -TL+TR cos(q)= 0 TR TL q TL = TR cos(q) y-direction: SF=ma W TR sin(q) – Mg = 0 Combine: TR = Mg / sin(q) TL = Mg cos(q)/sin(q) Physics 101: Lecture 3, Pg 4
  • 5. Overview Kinematics: Description of Motion Position and displacement velocity » average » instantaneous Acceleration » average » instantaneous Relative velocity (first pass) Physics 101: Lecture 3, Pg 5
  • 6. Position vs Time Plots  Gives location at any time.  Displacement is change in position.  Slope gives velocity. x (m) 3Position at t=3, x(3) = 1Displacement between t=5 and t=1. Dx = -1.0 m t 4 1.0 m - 2.0 m = -1.0 mAverage velocity between t=5 and t=1. v = -0.25 m/s -3 -1 m / 4 s = -0.25 m/s Physics 101: Lecture 3, Pg 6
  • 7. Velocity vs Time Plots v (m/s)  Gives velocity at any time.  Area gives displacement 3  Slope gives acceleration. 1.5 6 t velocity at t=2, v(2) = 3 m/s 4 Displacement between t=0 and t=3: Dx= 7.5 m -3 t=0 to t=1: ½ (3m/s) (1 s) = 1.5 m t=1 to t=3: (3m/s) (2 s) = 6 mAverage velocity between t=0 and t=3? v= 7.5 m / 3s = 2.5 m/sChange in v between t=5 and t=3. Dv = -2 m/s – 3 m/s = -5 m/sAverage acceleration between t=5 and t=3: a = -5 m/s / (2 s) = -2.5 m/s2 Physics 101: Lecture 3, Pg 7
  • 8. Acceleration vs Time Plots  Gives acceleration at any time.  Area gives change in velocity a (m/s2) 3Acceleration at t=4, a(4) = -2 m/s2 6Change in v between t=4 and t=1. Dv = +4 m/s t t=1-3: Dv = (3m/s2)(2s) = 6 m/s 24 t=3-4: Dv = (-2m/s2)(1s) = -2 m/s -3 Physics 101: Lecture 3, Pg 8
  • 9. Acceleration Preflights Is it possible for an object to have a positive velocity at the same time as it has a negative acceleration?88% 1 - Yes “the object could be slowing down.”12% 2 - No If the velocity of some object is not zero, can its acceleration ever be zero ?87% 1 - Yes “The velocity could be non-zero and constant. A constant velocity has no13% 2 - No acceleration..” Physics 101: Lecture 3, Pg 9
  • 10. Velocity ACTIf the average velocity of a car during a trip along a straightroad is positive, is it possible for the instantaneous velocity atsome time during the trip to be negative?A - YesB - NoDrive north 5 miles, put car in reverseand drive south 2 miles. Averagevelocity is positive. Physics 101: Lecture 3, Pg 10
  • 11. Dropped Ball y•A ball is dropped from a height of xtwo meters above the ground.  Draw vy vs t v v v9 A 9 B 9 C 0.5 t 0.5 t 0.5 t -6 -6 -6 v v 9 9 E D 0.5 t 0.5 t -6 -6 Physics 101: Lecture 3, Pg 11
  • 12. Dropped Ball xA ball is dropped for a height of twometers above the ground. t v  Draw v vs t  Draw x vs t t  Draw a vs t a t Physics 101: Lecture 3, Pg 12
  • 13. Tossed Ball•A ball is tossed from the ground up a height of twometers above the ground. And falls back down y x  Draw v vs t v v v 9 A 9 B 9 C 1 t 1 t 1 t -6 -6 -6 v v 9 9 D E 1 t 1 t -6 -6 Physics 101: Lecture 3, Pg 13
  • 14. Tossed Ball x•A ball is tossed from the ground up aheight of two meters above the ground. tAnd falls back down v  Draw v vs t  Draw x vs t t  Draw a vs t a t Physics 101: Lecture 3, Pg 14
  • 15. ACTA ball is thrown straight up in the air and returns to itsinitial position. During the time the ball is in the air,which of the following statements is true?A - Both average acceleration and average velocity are zero.B - Average acceleration is zero but average velocity is not zero.C - Average velocity is zero but average acceleration is not zero.D - Neither average acceleration nor average velocity are zero. Vave = DY/Dt = (Yf – Yi) / (tf – ti) = 0 aave = DV/Dt = (Vf – Vi) / (tf – ti) Not 0 since Vf and Vi are not the same ! 101: Lecture 3, Pg 15 Physics
  • 16. Relative Velocity (first pass)You are on a train traveling 40 mph North. If you walk 5 mph toward the front of the train, what is your speed relative to the ground?A) 45 mph B) 40 mph C) 35 mph 40 mph N + 5 mph N = 45 mph N 40 5 45 Physics 101: Lecture 3, Pg 16
  • 17. Relative VelocityYou are on a train traveling 40 mph North. If you walk 5 mph toward the rear of the train, what is your speed relative to the ground?A) 45 mph B) 40 mph C) 35 mph 40 mph N - 5 mph N = 35 mph N 40 5 35 Physics 101: Lecture 3, Pg 17
  • 18. Relative VelocityYou are on a train traveling 40 mph North. If you walk 5 mph sideways across the car, what is your speed relative to the ground?A) < 40 mph B) 40 mph C) >40 mph 40 mph N + 5 mph W = 41 mph N 5 | v | 40 2  5 2 40 Physics 101: Lecture 3, Pg 18
  • 19. Relative Velocity Sometimes your velocity is known relative to a reference frame that is moving relative to the earth. Example 1: A person moving relative to a train, which is moving relative to the ground. Example 2: a plane moving relative to air, which is then moving relative to the ground. These velocities are related by vector addition:   » vac is the velocity of the object relative to the groundv ac  v ab  v bc » vab is the velocity of the object relative to a moving reference frame » vbc is the velocity of the moving reference frame relative to the ground Physics 101: Lecture 3, Pg 19
  • 20. Tractor Demo 1Which direction should I point the tractor to get it across the table fastest?A) 30 degrees leftB) Straight acrossC) 30 degrees right 1 2 3 Physics 101: Lecture 3, Pg 20
  • 21. Tractor Demo (moving table) Which direction should I point the tractor to get it across the table fastest?A) 30 degrees leftB) Straight acrossC) 30 degrees right 1 2 3 Physics 101: Lecture 3, Pg 21
  • 22. Summary of Concepts kinematics: A description of motion position: your coordinates displacement: Dx = change of position velocity: rate of change of position average : Dx/Dt instantaneous: slope of x vs. t acceleration: rate of change of velocity average: Dv/Dt instantaneous: slope of v vs. t relative velocity: vac = vab + vbc Physics 101: Lecture 3, Pg 22

×