Boddeker's Mechanics Notes



 

|Chapter 2 Outline |

|2.1 Position, Velocity, & Speed (2.1 & 2.2 for 121) |2.5 1-D Motion w/Constant Accel |

|2.2 Instantaneous Velocity & Speed (2.3 for 121) |2.6 Freely Falling Objects |

|2.3 Acceleration |2.7 Kinematic Eq derived from Calculus |

|2.4 Motion Diagrams | |

|Position, Velocity, & Speed |

|Displacement – Distance from a given point, including direction from given point. |

|[pic] |

|+5 meters in the x-direction |+6 meters in the y-direction |-4 meters in the z-direction |The resultant 3D displacement |

|[pic] |[pic] |[pic] |[pic] |

|Note…this is not the typical 3D axial orientation, but a z-x-y orientation.  The z-x-y orientation is easier for students unfamiliar with 3D coordinates. |

|When you get to Linear Algebra or Multivariable calculus you’ll see the typical x-y-z orientation. |

|  |

|  |

|What’s the difference between speed and velocity? |

|  |

|What if you told me you drove for 1 hour from exit to exit on I-10, without delays, and ended up 65 miles from your starting point. |

|Could I guess how fast you were driving? |Could I know where you ended up? |

|  |  |

|Obviously…you covered 65 miles in one hour…thus your speed was 65 mph. |NO WAY…I don’t have any direction. |

|Speed is a scalar (just magnitude) quantity. |Velocity is speed WITH direction. |

|  |Velocity is a vector quantity. |

|Instantaneous Velocity & Speed  and   Acceleration |

|A speedometer supposedly gives instantaneous speed. |[pic] |

|Thus if you are starting from a stop, you notice the speedometer go faster and faster until you approach your desired maximum speed.  | |

|  | |

|Let’s assume you had a constant acceleration during this time; [pic] where acceleration is velocity versus time (a = Δv|Thus our position vs time graph looks like this [pic], |

|/ Δt) and is a straight line.  |where vave = Δx / Δt |

|  |

|Let’s assume our final speed after acceleration is about 65 mph. |

|So right after we started from rest, we know we are much less than 65 mph, probably less than 5 mph. |Near our maximum velocity a considerable time later, we should be close to 65 mph. |

|  |  |

|[pic] |[pic] |

|If you notice in both above sketches, the straight line touches the curve at a small area (theoretically at one point).  We now need to define a slope. |

|  |

|The slope is Δy / Δx, or in our case Δx / Δt, since displacement is on the y axis and time on the x axis.  |

|  |

|In the calculus based class, we say the slope is the derivative of a point, (slope = dy / dx).  |

|  |

|In the non-calculus based classes…we say the slope is the slope over a very small range about the point of interest |

|(in the first example, the point of interest was the slope at a very small time and |

|in the second example a large amount of time when we’ve almost achieved 65 mph.) |

|  |

|Demo: Dune Buggy 1-D:  ME-C-DB |

|  |

|2.4 Motion Diagrams |

|A.    Positive Velocity - your position progressed with time |B.    Negative Velocity – your position regresses with time |C.    Zero Velocity – your position remains constant with time |

|  |[pic] |[pic] |

|[pic] | | |

|D.    Positive Acceleration - your velocity increases with time |E.    Negative Acceleration – your velocity decreases with |F.    Zero Acceleration – your velocity remains constant with time |

|[pic] |time |[pic] |

| |[pic] | |

|G.    Positive Acceleration – your position progress as a square |This data represents which sketch? |Please review this section several more times before your midterm and final exam.|

|[pic] | |  |

| | |You can complete the negative acceleration and zero acceleration where position |

| | |is plotted version time. |

| |x(t) |time | |

| |(meters) |(seconds) | |

| |1 |1 | |

| |4 |2 | |

| |9 |3 | |

| |16 |4 | |

|  [pic]                 [pic]                    [pic] |

|2.5 1-D Motion w/Constant Accel |

|Demo: String and Weights Drop:  ME-C-SW |

|  |

|a = dv / dt; |

|we are assuming constant acceleration.   “During class we derive the equation when we assume Jerk is constant and                                                                                                |

|acceleration is varying” |

|The 1st lecture introduced the kinematics equations derived from calculus. |

|Integrated a = dv / dt twice. |

|x(t) = ½at² + vot + xo |

|Example 1 |What is the velocity after 5 seconds? |

|A car is driving in some odd traffic with the equation over a limited amount of time where |We will fall back upon the definition of instantaneous velocity |

|it position is give by the following equation: |v = dx / dt |

|  |v = ½ 2t + 3 |

|x(t) = ½t2 + 3t + 5 |v = 8 m/s |

|  | |

|What is the position of the car after 5 seconds? |What is the acceleration after 5 seconds? |

|x = ½52 + 3*5 + 5 |Once again we will use the definition of acceleration |

|x = 32.5 meters |a =    d2x / dt2 |

|  |a =     dv  / dt |

|Note:  Through out the rest of this course we ALWAYS assume acceleration is constant unless |a = d(3 + t)       / dt |

|otherwise indicated. |a = 1 m/s2 |

|No calculus method |What is the velocity after 5 seconds? |

|How far has the car traveled after |Using the definition: |

|5 seconds? |a = Δv / Δt |

|x = ½  a   t²      + vo t            + xo |a = (vf - vi) / t |

|x = ½ (1) 52       + 3*5    + 5 |1 = (vf - 3) / 5 |

|x = 32.5 meters |vf = 8 m/s |

|From the above comparison we know | |

|a  = 1 m/s2;   vo = 3 m/s;   xo = 5 m |What is the acceleration after 5 seconds |

| |Well...it never changes...it's 1 m/s² |

|  |  |

|Example 2: Bumper Cars |        [pic] |

|Red and blue bumper cars are heading toward each other. | |

|The blue one is traveling toward the red one at a constant velocity of 3.0 m/s.  | |

|The red car is initially at rest and accelerates at a rate of 1.0 m/s2 toward the | |

|blue car as the blue car is 10 meters away. | |

|a) When do the bumper cars meet? | |

|b) Where do the bumper cars meet? | |

|c) What is the velocity of the red car at this time? | |

| |(a)  As the two bumper cars meet, they’ll be at the same position |(b) |

| |x      = ½at²      + vot + xo |dred  = ½ 1 (2.39)2 |

| |xred   = ½1t2       +   0 + 0 |dred  = 2.84 m |

| |xred   = ½t2 |  |

| |xblue  = 0   +  -3t        + 10 |dblue = - 3*2.39   + 10 |

| |xblue  = -3t + 10 |dblue = 2.84 m |

| |  | |

| |        xred           =     xblue | |

| |        ½ t2          = -3t   + 10 | |

| |                t       = 2.4 sec | |

| | |(c)    a = Δv / Δt |

| | |        a = (vf - vi) / t |

| | |        1 = (vf - 0) / 2.4 |

| | |        vf = 2.4 m/s |

|Assumptions & Declarations |

|1. We will define the zero point of the system at the red bumper car. |

|2. Thus the blue bumper car has an initial position of 10 m. |

|3. We will define increase from the zero position as positive, thus the red car will have a positive velocity and acceleration. |

|4. The blue car is traveling toward the zero position, thus the velocity will be negative. |

|  |

|Example 3: Motorcycle Ramp |

|What length of road is required for a motorcycle to achieve 80 m/s (at the base of a |a = (vf - vi) / t |x = xo       + vo t        + ½  a   t2 |

|ramp) to jump a row of cars if its acceleration is 4 m/s2? |4 = (80 - 0) / t |x = 0        +  0          + ½  4  202     |

|  |t = 20 sec |x = 800 m |

| | |  |

|Example 4: Motorcycle Cop |You get a 2 second head start on the bike. |  |  |

|You are driving your car at its maximum speed of 134 mph (or about|xyou = ½at2   + vot          + xo    |xcop = ½at2        + vot           + xo    | |

|60 m/s).  As you are driving you pass a motorcycle police officer |xyou = 0       + 60 (t+2)   + 0 |xcop = ½20t2      + 0     + 0 | |

|and it take the officer 2 seconds to hop on his bike to start |xyou = 60t +120 |xcop = 10 t2 | |

|chasing you.  The motorcycle has a maximum velocity of 223 mph (or| | | |

|about 100 m/s).  The max rate of acceleration of the motorcycle is| | | |

|about 20 m/s2.  What distance does your car travel before being | | | |

|caught by the officer? | | | |

|  | | | |

| |When the bikes position equals your position then the bike has caught you. |  |

| |xyou           = xcop |But WAIT a second…this can’t be correct.  The bike hits maximum |  |

| |60t +120  = 10t2 |velocity at 5 seconds. | |

| |t2 – 6t              = 12 |        a      = Δv / t | |

| |t2 – 6t + 32       = 12 + 32 |        20    = (100 – 0)* t | |

| |(t – 3)2             = 21 |        t       = 5 sec | |

| |t = 7.58 seconds (can’t be negative) | | |

| |Start OVER !!! |  |

|Think About It |You get a 2 second head start on the bike. |  |  |

|For xcop why is the time, (t – 5), associated with the constant |xyou = ½at2 + vot        + xo    |  | |

|velocity? |xyou = 0     + 60 (t+2) + 0 |xcop = ½ a   t2    +   vo  t     + xo    | |

|Assume total time is 9 seconds…if the initial 5 seconds represents|xyou = 60t +120 |xcop = ½20 52    +100(t-5)  + 0 | |

|the acceleration, how much time is left for the bike to drive at |  |xcop = 100t - 250 | |

|top speed? | | | |

|          Obvious…4 seconds. | | | |

|Time remaining for constant velocity is represented by (t – 5) | | | |

|Time remaining = 9 – 5 = 4 sec. | | | |

| |  |        Test using both equations |  |

| |xyou         =      xcop | | |

| |60t +120 = 100t –250 | | |

| |t = 9.25 sec | | |

| |  | | |

| | |xyou = 60 (t+2) |xbike = ½20*52 + vf(t-5) |  |

| | |xyou = 675 m |xbike = 675 m | |

|  |  |

|Example 5: Arrow Impacting Tree | |

|An arrow is shot in a tree.  The arrow penetrates 10 cm into the tree.  The rate of acceleration is -40,000 m/s2.  How fast was the arrow traveling before hitting tree? | |

| (Remember…the negative sign just states it is slowing down) |(Rigorous Method) |Δv = vf – vi       ⋄ vf = 0 m/s |  |

|x      = ½     a           t2 |a = Δv / Δt |-vi = a*t | |

|0.1 m        = ½ 40,000 t2   |  |-vi = -40000t | |

|t = 0.00224 sec | | | |

|  | | | |

|a        = Δv           /     Δt | | | |

|-40,000 = (0 – vi) / 0.00224 s | | | |

|vi = 90 m/s | | | |

|which is approximately 200 mph | | | |

| |0 = ½-40000t2 + vo         * t              + 0.1    |  |

| |0 = ½-40000t2 + -40000t  * t          + 0.1  | |

| |0 = -20000t2    – 40000t2                 + 0.1 | |

| |20000t2 = 0.1 | |

| |                t = 0.00224 sec | |

| |(Same for rest of problem) | |

|As you can see the BELOW calculus method brings you to the Rigorous (Formal) method |  |

|a = dv / dt |v = dx / dt |  |

|-40,000 dt = dv |(-40,000t + vo)dt = dx | |

|-40,000 ∫ dt = ∫ dv |-40,000 ∫ tdt + vo∫dt = ∫ dx | |

|v = -40,000 t + vo |x = -20,000 t2 + vot + xo   | |

| |  | |

|Example: |Ans:  (assume “g” = 10 m/s2) |"How do you know?" your friend asks. |  |

|Dropping a rock down a well? |You decide to put you newly acquired kinematics skills to |You explain what you did. | |

|After a very dry spell, you decide to measure the depth of your |work. |Your friends responds, "Woah, we really have a little more water left than you | |

|well.  Normally the water level is 12 meters below the surface of |You carefully time a dropped rock down your well and measure |calculated, let me show you." | |

|the well.  You also know the well was initially dug to 45 meters, |an echo 3 seconds later |You've neglected the amount time for sound to travel back to the top of the well, | |

|so you have an ample supply of well water during dry climate.  The|d = ½ a t² |where vsound = 300 m/s | |

|only instrument you have is your watch with a timer on it. How can|d = ½ 10 3² |We've forgot to mention...you live in North Dakota and it's January!!! | |

|you determine the depth of the well? |d = 45 m |At STP, vsound = 343 m/s | |

| |  | | |

| |You run to your roomate (also in your physics class) and | | |

| |say...we're almost out of water!!! | | |

|Think About It |dsound = vsound * tsound |v(3) – v(trock)     =  ½a trock² |  |

|We know that ttotal is comprised of both trock and tsound such |dsound = vsound * (3 - trock) |900 - 300t       = 5 t² | |

|that |  |t² + 60t  = 180 | |

|ttotal          = trock   + tsound |drock  = ½ atrock² |(t + 30)² = 180 + 30² | |

|3 sec        = trock   + tsound |But both distances are equal; so |t               = -30 ± + 32.86 | |

|  |  |trock = 2.8634 | |

|Solve for tsound in terms of trock |dsound                                    =   drock |tsound = 3 - 2.8634 | |

|  |vsound (3-trock)    = ½ atrock2 |tsound = 0.1365 seconds | |

|tsound = 3 - trock | | | |

|vy = Δy / Δt;  where Δy = d | | | |

|  | | | |

|dsound = vsound * tsound | | | |

|Test  ⋄ |dsound = vsound * tsound |drock    = ½ atrock² |  |

| |dsound = 300 * 0.1365 |drock    = ½ 10 * 2.8634² | |

| |dsound = 41 m |drock    = 41 m | |

|2.6 Freely Falling Objects |  |

|[pic]   |  |

|As soon as an object leaves a persons hand, exits the|Example 1: Dropped Watermelon meets Cannon Ball |[pic] |  |

|barrel of the cannon, engines shut off, etc the |Both the watermelon and the cannon ball are released/fired at the same time. | | |

|object ONLY has one force acting on it which is |The watermelon is dropped from a building of height, H. | | |

|gravity.  This is called free fall.  |The cannon is at the bottom of the building. | | |

|  |  | | |

|The acceleration due to gravity is 9.81 m/s2. |The cannon impacts with the watermelon at 0.6 H (from the ground). | | |

|Many times “g” is estimated to 9.8 m/s2 or even 10 |a) How much time has elapsed since the two objects were released? | | |

|m/s2. |b) What is the initial velocity of the cannonball? | | |

| |  | | |

| |This problem will be solved using two methods. | | |

| |Point of View perspective and the Rigorous Method | | |

|Point of View Perspective |Rigorous Method |  |

|  |Here we define the bottom is 0 meters and the top is H. | |

|The cannon ball travels a total distance of 3H/5 where the watermelon travels the remaining |In the Rigorous Method acceleration due to gravity is ALWAYS negative. | |

|distance (2H/5), where the total distance is H. |The watermelon will fall to the position 3H/5 and the cannon ball will rise to 3H/5. Thus the equations will be | |

|  |Watermelon | |

|In Point of View method,  velocity is increasing as it goes down, thus a positive 10 m/s2 |Δy    = ½  ay   t2 + vot + yo | |

|acceleration. |3H/5        = ½(-10) t2        +  0  + H     | |

| |t       = ( 2h / 25 )½ | |

|Watermelon |Note: this is the same conclusion as the Point of View Perspective | |

|Δy    = ½  ay t2   + vot         + yo | | |

|2H/5        = ½(10)t2   +  0          + 0 |Cannon Ball | |

|t       = √(2H/25) |Δy     = ½ayt2            + vot + yo | |

|  |3H/5 = ½(-10) t2       + vo t        + 0 | |

|Cannon Ball |        where t = √(2H/25) | |

|Δy    = ½   ay t2  + vot + yo |3H/5 = ½(-10)(2H/25)      + vo ( 2H / 25 )½ + 0 | |

|3H/5        = ½(-10)t2 + vot + 0 |3H/5 = -2H/5           + vo ( 2H / 25 )½ | |

|        where t = √(2H/25) |H = vo ( 2H )½ / 5 | |

|3H/5        = ½(-10)2H/25  + vo √(2H/25) |vo = (25H/2)½ | |

|H = vo √(2H) / 5 | | |

|vo = √(25H/2) | | |

|2.7 Kinematic Eq derived from Calculus |  |

|If you want to find the maximum position of the position equation…how do you do it? |  |

|Rules from Calculus | |

|Take the derivative of your given equation and set it equal to zero. | |

|Solve for time, and this is the time that you have your maximum position. | |

|x(t)  = 5t2 - 20t  +   16 | |

|0 = 10t – 20 | |

|t = 2 sec | |

|x(2) = 5t2 - 20t  +   16 | |

|x(2) = -20 meters | |

|  | |

|From INTRO notes |  |  |  |

|  |a = Δv / Δt |  | |

|v = dx / dt |and instantaneous acceleration is |Once more: | |

| | [pic]  eq 2 |Integrate with respect to time | |

|[pic]   eq 1 | |  | |

| | |dx /dt = at + vo | |

| | |  | |

| | |∫ dx = a ∫ tdt   +  vo ∫ dt | |

| | |  | |

| | |x = ½ at2 + vot + xo     eq 3 | |

| | |  | |

|I prefer you ONLY using the following three formulas for ALL kinematics problems |  |

|  |  |  |  |

|1)   [pic] |2)   [pic] |3)   x = ½ at2 + vot + xo | |

| | | | | | | | | | | | | | | | | | | | | | | | | | | | | 

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