Lecture04
- 1. Kinematics + Dynamics Today’s lecture will cover Textbook Chapter 4 iclicker scores have been imported again – please check! If you are new to the course, please read the course description on the course web page! Physics 101: Lecture 04 Exam I neptune
- 2. What’s Most Difficult Tension / setting up problem / vectors / free body diagrams. … which problems out of the book would be most beneficial to use for practice.. Knowing if we must memorize te equations . remembering to do the prefilghts / Staying awake. Trig!! SOH CAH TOA?? 6 years of information sort of nudged that out of my brain getting used to symbolic representations of equations
- 3. Review Kinematics : Description of Motion Position Displacement Velocity v = x / t average instantaneous Acceleration a = v / t average instantaneous Relative velocity: v ac = v ab + v bc
- 4. Preflight 4.1 (A) (B) (C) Which x vs t plot shows positive acceleration? 88% got this correct!!!! “ B has positive acceleration because it has more of an angled curve that is concave up. Plus, it's red, and red things move faster ” … interpreting graphs…
- 5. Equations for Constant Acceleration (text, page 113-114) x = x 0 + v 0 t + 1/2 at 2 v = v 0 + at v 2 = v 0 2 + 2a(x-x 0 ) x = v 0 t + 1/2 at 2 v = at v 2 = v 0 2 + 2a x 14
- 6. Kinematics Example A car is traveling 30 m/s and applies its breaks to stop after a distance of 150 m. How fast is the car going after it has traveled ½ the distance (75 meters) ? A) v < 15 m/s B) v = 15 m/s C) v > 15 m/s This tells us v 2 proportional to x
- 7. Acceleration ACT A car accelerates uniformly from rest. If it travels a distance D in time t then how far will it travel in a time 2t ? A. D/4 B. D/2 C. D D. 2D E. 4D Follow up question: If the car has speed v at time t then what is the speed at time 2t ? A. v/4 B. v/2 C. v D. 2v E. 4v Demo… Correct x=1/2 at 2 Correct v=at
- 8. Newton’s Second Law F=ma position and velocity depend on history Net Force determines acceleration
- 9. ACT A force F acting on a mass m 1 results in an acceleration a 1 . The same force acting on a different mass m 2 results in an acceleration a 2 = 2 a 1 . What is the mass m 2 ? (A) 2m 1 (B) m 1 (C) 1/2 m 1 F=ma F= m 1 a 1 = m 2 a 2 = m 2 (2a 1 ) Therefore, m 2 = m 1 /2 Or in words… twice the acceleration means half the mass F a 1 m 1 F a 2 = 2a 1 m 2
- 10. Example: A tractor T (m=300Kg) is pulling a trailer M (m=400Kg). It starts from rest and pulls with constant force such that there is a positive acceleration of 1.5 m/s 2 . Calculate the horizontal force on the tractor due to the ground. X direction: Tractor F = ma F w – T = m tractor a F w = T+ m tractor a X direction: Trailer F = ma T = m trailer a Combine: F w = m trailer a + m tractor a F w = (m trailer+ m tractor ) a F W = 1050 N y x T W N T F w W N
- 11. Net Force ACT Compare F tractor the net force on the tractor, with F trailer the net force on the trailer from the previous problem. A) F tractor > F trailor B) F tractor = F trailor C) F tractor < F trailor F = m a F tractor = m tractor a = (300 kg) (1.5 m/s 2 ) = 450 N F trailer = m trailer a = (400 kg) (1.5 m/s2) = 600 N
- 12. Pulley Example Two boxes are connected by a string over a frictionless pulley. Box 1 has mass 1.5 kg, box 2 has a mass of 2.5 kg. Box 2 starts from rest 0.8 meters above the table, how long does it take to hit the table. 1 2 1) T - m 1 g = m 1 a 1 2) T – m 2 g = -m 2 a 1 2) T = m 2 g -m 2 a 1 1) m 2 g -m 2 a 1 - m 1 g = m 1 a 1 a 1 = (m 2 – m 1 )g / (m 1 +m 2 ) Compare the acceleration of boxes 1 and 2 A) |a 1 | > |a 2 | B) |a 1 | = |a 2 | C) |a 1 | < |a 2 | 1 T m 1 g 2 T m 2 g y x
- 13. Pulley Example Two boxes are connected by a string over a frictionless pulley. Box 1 has mass 1.5 kg, box 2 has a mass of 2.5 kg. Box 2 starts from rest 0.8 meters above the table, how long does it take to hit the table. 1 2 a 1 = (m 2 – m 1 )g / (m 1 +m 2 ) a = 2.45 m/s 2 x = v 0 t + ½ a t 2 x = ½ a t 2 t = sqrt(2 x/a) t = 0.81 seconds Compare the acceleration of boxes 1 and 2 A) |a 1 | > |a 2 | B) |a 1 | = |a 2 | C) |a 1 | < |a 2 | 1 T m 1 g 2 T m 2 g y x
- 14. Summary of Concepts Constant Acceleration x = x 0 + v 0 t + 1/2 at 2 v = v 0 + at v 2 = v 0 2 + 2a(x-x 0 ) F = m a Draw Free Body Diagram Write down equations Solve Next time: textbook section 4.3, 4.5