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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