Note_02 1D motion PHYS 204-04 Mechanics PDF

Note_02 Chap. 2 Motion in 1-dimension PHYS 204-04 Mechanics January 2025 Definition () of distance, displacement, speed and velocity.  Location or position requires a frame of reference, typically some xy coordinate system. For example, you are at the int...

Note_02 Chap. 2 Motion in 1-dimension PHYS 204-04 Mechanics January 2025 Definition () of distance, displacement, speed and velocity.  Location or position requires a frame of reference, typically some xy coordinate system. For example, you are at the intersection of abc road and the cross street is cde road.  Distance  Displacement.  Distance (understood to mean traveled, a scalar quantity)  total distance of traveled regardless of direction. Travel from Montreal to Toronto and back, the distance is approximately 500km  2. Here we choose Montreal as the starting point. North 500 km Finish distance = 1000 km Start 500 km East Toronto Montreal   Displacement (directional, a vector quantity)   x  final position less initial position.  The journey is divided into many segments x due to rest room stops and gas up, and may be off the main highway. [An over score indicates a vector quantity. Detail will be treated in Chapter 3] The displacement of traveling from Montreal to Toronto and back, is zero km. North 500 km Finish displacement =-500+500=0 km Start 500 km East Toronto Montreal total distance travelled  Speed (scalar, always implies average over time) vavg . total time taken   total displacement  x (vector)  Velocity (average and directional) v  . total time  t  The displacement can have many segments x with corresponding time t.    lim x dx In the limit t dt ,  Instantaneous velocity (directional) v  .    t 0 t dt x dx The same symbols are used for both speed and velocity, but with added qualifications. Page 1 of 5 Note_02 Chap. 2 Motion in 1-dimension PHYS 204-04 Mechanics January 2025 Problem 2-4 p.48 of text book. An athlete leaves one end of a pool of length L at t  0 and arrives at the other end at time t1. She swims back and arrives the starting position at time t2. If she is swimming initially in the positive x direction, determine her average velocities symbolically in (a) the first half of the swim, (b) the second half of the swim, and (c) the round trip. (d) What is her average speed for the round trip? Answer It is best to represent the given info in terms of a diagram. y this dimension not used here. pool length L Start x Finish  L (a) 1st lap: vave  , default direction to the right in diagram is positive. t1  L (b) 2nd lap: vave  , negative means to the left. t2  displacement  L  L (c) round trip: vave    0. Start and finish are at the same location. time t1  t2 total distance | x1 |  | x2 | L  L (d) round trip: speed ave    0 total time t1  t2 t1  t2 The different meaning of speed and velocity is demonstrated in part (c) & (d). For individual lap, speed and velocity have the same numerical value. Acceleration  Speed and velocity are usually a function of time.  Acceleration  time rate of change in speed (more details: direction, increase of decrease in time).  In nearly all situations in PHYS 204 the acceleration is assumed to be constant in time. The only exception is when dealing with the spring-mass problems.  In order to have an acceleration, a force (contact or non-contact) must be present. (Chapter 5 will deal force with force. Newton’s 2nd law: a   constant for a constant force. ) mass Description of acceleration as a functdion of t. Graphical representation.  Acceleration as a function of time t. |acceleration vs t| Diagram at right → rather uneventful, as it is constant in t. a t Page 2 of 5 Note_02 Chap. 2 Motion in 1-dimension PHYS 204-04 Mechanics January 2025  Connection between v and constant acceleration. v(t) For constant acceleration, the velocity v increases linearly with time t. v Can also be decreasing linearly with time. v(t)-vi vi t change of v v t By definition, a   = slope of v(t ) vs t. t’ change of t t t From the right triangle (blue and dotted lines) it can be seen, v(t )  vi a  v(t )  vi  at , a useful formula for constant a.............................................(2-13) t v  vi  a t , a straight line function, slope  a, y-intercept  vi. a good formula to remember.  Connection between v and x, both are function of time t. v x At any time t , by definition, v(t )  in the neighborhood of t . v(t)-vi t v(t’) t’ t x  v(t )  t  area of a the cross-hatched panel, near t . t t Total displacement x  xi   v(t )  t   many vertical panels or area under the function v vs t. From the diagram of v(t ) vs t , the area = rectangle + triangle,   x  xi   v(t )  t  vi t     vi t  t   vi  at    vi   vi t  12 a t 2 1 1 t  v(t )  vi   2   2   rectangle triangle  due to accel  Given (vi , t , and a ) , x  xi can be calculate with the above standard formula. 1 x  xi  vi t  a t 2 , a useful formula for constant a.....................................................(2-16) 2  For constant a (through out the course), and if we are given (vi , v f , and a) how to find ( x  xi ) ? v f  vi v f  vi a  aavg   t t a x  xi  vi v f  vi 1  a (v f  vi ) 2   ( 2vi v f  2vi2 )  v 2f  2v f vi  vi2  v 2 f  vi2............(2-17) 2 2 a  a  2a 2a t t2 At least one of the formulae (2-13), (2-16) and (2-17) can be used to solve problems in 1-dimension motion. The choice depends on what parameters are given and what unknown is sought. Train of thought: Start with constant acceleration. Derive velocity from acceleration, then derive displacement from velocity. “Derive” means using arguments based on definitions or rules. Do not do it in the reverse order. Page 3 of 5 Note_02 Chap. 2 Motion in 1-dimension PHYS 204-04 Mechanics January 2025 Problem 2-42, p.51 of textbook. [only part (a) is quoted below] Two thin rods are fastened to the inside of a circular ring as shown in Figure P2.42. A One rod of length D is vertical, and the other of length L makes an angle  with the horizontal. The two rods and the rings are in the vertical plane. Two small beads B D are free to slide without friction along the rods. (a) If the two beads are released L from rest simultaneously form the position shown, use your intuition and guess which bead reaches the bottom first.  C Answer: Figure P2.42 Any clue where to start? This is a typical problem that makes people think physics is a difficult subject. The difficulty lies in the connection of information from the question and some of the formulae developed for the 1-dimension motion. (1) Avoid the approach of looking for a formula. Digest the A question first.  We know the beads are confined to a straight line so R  B 1 R   " x  a t 2 " can be adapted to the problem. Making a good O ar=g 2 aL= gsin  L sketch that shows the given info will definitely help. R  g Visual cues are useful.  C (2) Read the question a few times and identify which symbol represents the unknown. (3) Working backward and forward to connect what is sought and what is given, and be very patient. 1 2 4R Red bead: 2 R  gtr  tr2 . tr is the time for the red bead to reach the point C. 2 g 1 2L Blue bead: L  ( g sin  )tb2  tb2 . (finding the acceleration along L needs Chap. 3). 2 g sin  In order to compare tr and tb we need the relationship between R, L and  Observe the two isosceles triangles AOB and OCB. Label the base angles as  and . This part is not intuitive, but will be found by staring at the problem long enough. The total interior angle of ACB  (2  2  )     (   )  90. Know your trig. L Since  is the complement of  and  is the complement of .  sin   sin   2R 2L 2(2 R sin  ) 4 R exactly tb2     tr2 g sin  g sin  g Final answer: the two beads arrive at the bottom at exactly the same time. Page 4 of 5 Note_02 Chap. 2 Motion in 1-dimension PHYS 204-04 Mechanics January 2025 An example of adapting the 1-dimension motion to satisfy a set of conditions. There are information that you have to deduce from the wording but are not spelled out. Example 2.10, p.43 of textbook Not a Bad Throw for a Rookie! A stone thrown from the top of a building is given an initial velocity of 20.0 m/s straight upward. The stone is launched 50.0 m above the ground, and the stone just misses the edge of the roof on its way down as shown in Figure 2.14. (a) Using t A  0 as the time the stone leaves the thrower’s hand at position A, B determine the time at which the stone reaches its maximum height. (b) Find the maximum height of the stone. (c) Determine the velocity of the stone when it returns to the height from which it A was thrown. Answer: (a)The equations describing the motion upward of point A are: v(t )  vi  at  vi  gt   1 2  y (t )  50  vit  2 gt  formula (2-16) building We would like to get the time tB from solving these two equations. 50.0m We can deduce at point B, the stone comes to a stop. 20  0  20  gt B  tB   2.04 s 9.8 (b) At point B, use the same formula (2-16): 1 2 9.8(2.04) 2 y  50  vi t  gt  50  20(2.04)  2 2 Figure 2.14  50  40.8  20.4  70.4 m above ground. (c) Because the distance traveled is the same (deceleration up and acceleration down), it can be deduced the time of flight from B to A must be the same as from A to B. " v  vi  a t " " v  vi  a t "  On the way up: . On the way down:  20 0  20  gt B vB  0  gt B  0  g g  20 m/s  Page 5 of 5