REVIEW Phys-151

1. REVIEWPhys-151 • Lectures 1-11 • Displacement, velocity, acceleration • Vectors • Newton’s laws of motion • Friction, pulleys

2. See text : 1-3 Atomic Density • In dealing with macroscopic numbers of atoms (and similar small particles) we often use a convenient quantity called Avogadro’s Number, NA = 6.02 x 1023. • Molar Mass and Atomic Mass are nearly equal • 1. Molar Mass = mass in grams of one mole of the substance. • 2. Atomic Mass = mass in u (a.m.u.) of one atom of a substance, is approximately the number of protons and neutrons in one atom of that substance. • Molar Mass and Atomic Mass are other units for density. What is the mass of a single carbon atom ? = 2 x 10-23 g/atom

3. x t v t a t Displacement, Velocity, Acceleration • If the position x is known as a function of time, then we can find both velocity vand acceleration a as a function of time!

4. Since ais constant, we can integrate this using the above rule to find: • Similarly, since we can integrate again to get: 1-D Motion with constant acceleration • High-school calculus: • Also recall that

5. Solving for t: Derivation: • Plugging in for t:

6. Alice v0 Bill v0 H vA vB Lecture 2, ACT 4 Alice and Bill are standing at the top of a cliff of heightH. Both throw a ball with initial speedv0, Alice straightdownand Bill straightup. The speed of the balls when they hit the ground arevAandvBrespectively. Which of the following is true: (a) vA < vB(b) vA = vB (c) vA > vB

7. y rx = x = r cos ry = y = r sin (x,y) r ry  x rx arctan( y / x ) See text: 3-1 Converting Coordinate Systems • In circular coordinates the vector R = (r,q) • In Cartesian the vector R = (rx,ry) = (x,y) • We can convert between the two as follows: • In cylindrical coordinates, r is the same as the magnitude of the vector

8. U û • Useful examples are the cartesian unit vectors [i, j, k] • point in the direction of the x, yand z axes. R = rxi + ryj + rzk y j x i k z See text: 3-4 Unit Vectors: • A Unit Vector is a vector having length 1 and no units. • It is used to specify a direction. • Unit vector u points in the direction of U. • Often denoted with a “hat”: u = û

9. AB = |A| |B| cos(q) q AB = 0 AB = 0 q |AB| = |A| |B| sin(q) AB = 0 AB = 0 Multiplication of vectors / Recap • There are two common ways to multiply vectors • “scalar product”: result is a scalar • “vector product”: result is a vector

10. v h  yo D Problem 1: • We need to find how high the ball is at a distance of 113m away from where it starts. Animation

11. Problem 1 • Variables • vo = 36.5 m/s • yo = 1 m • h = 3 m • qo = 30º • D = 113 m • a = (0,ay)  ay = -g • t = unknown, • Yf – height of ball when x=113m, unknown, our target

12. y v v0y  y0 v0x g x Problem 1 • For projectile motion, • Equations of motion are: vx = v0xvy = v0y - g t x = vx t y = y0 + v0y t - 1/ 2 g t2 And, use geometry to find vox and voy Find v0x = |v| cos . and v0y = |v| sin .

13. t = 0 TARGET y v0  t = 0 x PROJECTILE Problem 3 (correlated motion of 2 objects in 3-D) • Suppose a projectile is aimed at a target at rest somewhere above the ground as shown in Fig. below. At the same time that the projectile leaves the cannon the target falls toward ground. Would the projectile now miss or hit the target ? t = t1

14. y v (x,y) R x a = 0 a a = const. See text: 4-4 What is Uniform Circular Motion (UCM) ? • Motion in a circle with: • Constant Radius R • Constant Speed v =|v| • acceleration ?

15. v = R Polar Coordinates... • In Cartesian co-ordinates we say velocity dx/dt = v. • x = vt • In polar coordinates, angular velocity d/dt = . •  = t •  has units of radians/second. • Displacement s = vt. but s = R = Rt, so: y v R s t x

16. Lecture 5, ACT 2Uniform Circular Motion • A fighter pilot flying in a circular turn will pass out if the centripetal acceleration he experiences is more than about 9 times the acceleration of gravity g. If his F18 is moving with a speed of 300 m/s, what is the approximate diameter of the tightest turn this pilot can make and survive to tell about it ?

17. v Similar triangles: v1 v2 v2 R So: v1 R Acceleration in UCM: • This is called Centripetal Acceleration. • Now let’s calculate the magnitude: But R = vt for smallt

18. Lecture 6,ACT 2Uniform Circular Motion A satellite is in a circular orbit 600 km above the Earth’s surface. The acceleration of gravity is 8.21 m/s2 at this altitude. The radius of the Earth is 6400 km. Determine the speed of the satellite, and the time to complete one orbit around the Earth. • Answer: • 7,580 m/s • 5,800 s

19. Lecture 6,ACT 3Uniform Circular Motion A stunt pilot performs a circular dive of radius 800 m. At the bottom of the dive (point B in the figure) the pilot has a speed of 200 m/s which at that instant is increasing at a rate of 20 m/s2. What acceleration does the pilot have at point B ?

20. See text: Chapter 5 Dynamics • Isaac Newton (1643 - 1727) proposed three “laws” of motion: Law 1: An object subject to no external forces is at rest or moves with a constant velocity if viewed from an inertial reference frame. Law 2: For any object, FNET = F = ma Law 3: Forces occur in pairs: FA ,B = - FB ,A (For every action there is an equal and opposite reaction.)

21. See text: 5-4 Newton’s Second Law The acceleration of an object is directly proportional to the net force acting upon it. The constant of proportionality is the mass. • Units • The units of force are kg m/s2 = Newtons (N) • The English unit of force is Pounds (lbs)

22. Lecture 7,ACT 1Newton’s Second Law I push with a force of 2 Newtons on a cart that is initially at rest on an air table with no air. I push for a second. Because there is no air, the cart stops after I finish pushing. It has traveled a certain distance (before removing the force). F= 2N Cart AirTrack For a second shot, I push just as hard but keep pushing for 2 seconds. The distance the cart moves the second time versus the first is (before removing the force) : A) 8 x as long B) 4 x as long C) Same D) 2 as long E) can’t determine

23. Cart Cart Dx1 Dx2 Lecture 7, ACT 1 t2 =2s, v2 t1 =1s, v1 to , vo = 0 F= 2N Cart AirTrack A) 8 x as long B) 4 x as long C) Same D) 2 as long E) can’t determine B) 4 x as long

24. Cart Lecture 7, ACT 1a What is the distances traveled after Fapp removed in the two cases: (i) after applying Fapp for 1 s vs. (ii) after aplying Fapp for 2 s ? Fapp Cart Cart at rest AirTrack A) 8 x as long B) 4 x as long C) Same D) 2 as long E) can’t determine

25. otherwise v1=v01, cart keeps moving ! Cart Cart Dx1 Dx2 Lecture 7, ACT 1aWhat is the distances traveled after Fapp removed ? Ftot = 0 ? Fapp = 0 t1 , v1 = 0 to , vo1 Fapp= 2N Cart Cart at rest AirTrack Fapp = 0 Ftot = 0 ? to , vo2 t2 , v2 = 0 Fapp= 2N Cart Cart AirTrack at rest B) 4 x as long

26. Lecture 8, Act 2 • You are going to pull two blocks (mA=4 kg and mB=6 kg) at constant acceleration (a= 2.5 m/s2) on a horizontal frictionless floor, as shown below. The rope connecting the two blocks can stand tension of only 9.0 N. Would the rope break ? • (A) YES (B) CAN’T TELL (C) NO a= 2.5 m/s2 rope A B

27. total mass ! Fapp a= 2.5 m/s2 mB mA T mA T -T rope mB Fapp -T Lecture 8, Act 2Solution: • What are the relevant forces ? Fapp = a (mA + mB) Fapp = 2.5m/s2( 4kg+6kg) = 25 N T = a mA T = 2.5m/s2 4kg = 10 N T > 9 N, rope will brake aA = a = 2.5 m/s2 ANSWER (A) a = 2.5 m/s2 Fapp - T = a mB T = 25N - 2.5m/s2 6kg=10N T > 9 N, rope will brake a = 2.5 m/s2 THE SAME ANSWER -> (A)

28. j ma mg sin  i N mg cos  mg See text: Example 5.7 Inclined plane... • Consider x and y components separately: • i: mg sin = maa = g sin  • j: N - mg cos = 0. N = mg cos  m  

29. T2 T2 Eat at Bob’s y T1 q2 q1 x mg Free Body Diagram T1 q2 q1 mg Add vectors : Vertical (y): mg = T1sinq1 + T2sinq2 Horizontal (x) : T1cosq1 = T2cosq2

30. T1 T2 M2 M1 a Example-1 with pulley • Two masses M1 and M2are connected by a rope over the pulley as shown. • Assume the pulley is massless and frictionless. • Assume the rope massless. • If M1 > M2 find : • Acceleration of M1 ? • Acceleration of M2 ? • Tension on the rope ? Video Animation Free-body diagram for each object

31. T4 T1 T3 T2 F T5 < M Example-2 with pulley • A mass M is held in place by a force F. Find the tension in each segment of the rope and the magnitude of F. • Assume the pulleys massless and frictionless. • Assume the rope massless. • We use the 5 step method. • Draw a picture: what are we looking for ? • What physics idea are applicable ? Draw a diagram and list known and unknown variables. Newton’s 2nd law : F=ma Free-body diagram for each object

32. T4 T3 F=T1 T4 T2 T1 T3 T2 T2 T3 F T5 T5 < M M T5 Mg Pulleys: continued • FBD for all objects

33. T4 T3 F=T1 T2 T2 T3 T5 M T5 Mg Pulleys: finally • Step 3: Plan the solution (what are the relevant equations) • F=ma , static (no acceleration: mass is held in place) T5=Mg T1+T2+T3=T4 F=T1 T2+T3=T5

34. F=T1 T5=Mg T2+T3=T5 T1+T2+T3=T4 T4 T1=T3 T2=T3 T1 T3 T2 F T5 < T1=T2=T3=Mg/2 and T4=3Mg/2 M T5=Mg and F=T1=Mg/2 Pulleys: really finally! • Step 4: execute the plan (solve in terms of variables) • We have (from FBD): • Pulleys are massless and frictionless T2+T3=T5gives T5=2T2=Mg T2=Mg/2 • Step 5: evaluate the answer (here, dimensions are OK and no numerical values)

35. See text: 6-1 Force of friction acts to oppose motion: • Dynamics: i :F KN = m a j :N = mg so FKmg=m a j N F i ma K mg mg

36. Lecture 9, ACT 4 In a game of shuffleboard (played on a horizontal surface), a puck is given an initial speed of 6.0 m/s. It slides a distance of 9.0 m before coming to rest. What is the coefficient of kinetic friction between the puck and the surface ? A. 0.20 B. 0.18 C. 0.15 D. 0.13 E. 0.27

37. m2 T1 m1 m3 ExampleProblem 5.40 from the book Three blocks are connected on the table as shown. The table has a coefficient of kinetic friction of 0.350, the masses are m1 = 4.00 kg, m2 = 1.00kg and m3 = 2.00kg. What is the magnitude and direction of acceleration on the three blocks ? What is the tension on the two cords ?

38. T12 T23 T12 m2 a T23 T1 m2 mk m2g a a m1 m2g m1g m3g m3 m1 m3 T12 T23 T12 T23 T23 - m3g = m3a T12 - m1g = - m1a -T12 + T23 + mk m2g = - m2a SOLUTION: T12 = = 30.0 N , T23 = 24.2 N , a = 2.31 m/s2 left for m2

39. An example before we considered a race car going around a curve on a flat track. N Ff mg What’s differs on a banked curve ?

40. ExampleGravity, Normal Forces etc. Consider a women on a swing: Animation When is the tension on the rope largest ? Is it : A) greater than B) the same as C) less than the force due to gravity acting on the woman (neglect the weight of the swing)