Boddeker's Mechanics Notes
|Chapter 4: Motion in Two Dimensions |
| 4.1 Position, Velocity, Acceleration Vectors |4.4 Uniform Circular Motion |
|4.2 2D Motion w/Constant Acceleration |4.5 Tangential Radial Acceleration |
|4.3 Projectile Motion |4.6 Relative Velocity and Acceleration |
|4.1 Position, Velocity, Acceleration Vectors |
|[pic] |
|4.2 Motion w/Constant Acceleration |
|An Olympic swimmer crosses a river by attempting to swim straight |[pic] |or [pic] |[pic] [pic] |
|across. A 1 m/s current is pushing the swimmer downstream as the | | | |
|swimmer crosses the 10 meter river. | | | |
|a) How much time is required for the swimmer to cross? | | | |
|b) What is the final position of the swimmer on the other side? | | | |
|Write the known’s | | | |
|vy = 3 m/s | | | |
|vx = 1 m/s | | | |
|Δy = 10 meters | | | |
| | |b) [pic] |r2 = rx2 + ry2 |
| |a) |1 m/s = Δx/3.33 |r2 = 3.332 + 102 |
| |3 m/s = 10m/Δt |Δx = 3 1/3 m |r = 10.5 meters |
| |Δt = 3 1/3 sec. | |tan θ = rx / ry |
| | | |tan θ = 3.33 / 10 |
| | | |θ = 18.4° |
| |Later in the term we get to write the resultant position as r = 3.33 i + 10 j | |
| |But until we complete our Mid-Term we calculate the magnitude of the resultant |10.5 meters & 18.4° downstream |
| |with angle. |(3.33 meters downstream) |
|This time the swimmer |vy = v (cos θ); |[pic] |[pic] |b) [pic] |
|desires to go directly | | | |v (cos θ) = Δy / Δt |
|across the river and compensate for the current. |vx = v (sin θ); | | |3 (cos θ) = 10 / Δt |
|a) Which direction should | | | |Δt = 3.33 / cos θ |
|the swimmer point so that | | | | |
|the swimmer will reach the other side directly across from the start| | | |Δt = 3.33 / cos 19.5° |
|point? | | | |Δt =3.54 seconds |
|b) How much time is required? | | | | |
|[pic] | | | | |
| |If Δx is zero then vx of the swimmer must equal vx of the | | |
| |river | | |
| | | |
| |a) vx-swimmer = vx-river | |
| |v (sin θ) = 1 m/s | |
| |3 (sin θ) = 1 m/s | |
| |θ = 19.5° upstream | |
|4.3 Projectile Motion |
|Key points to emphasize |
|Always break a vector into its horizontal and vertical (x & y) components IMMEDIATELY! |
|x & y components are independent |
|gravity (“g” = 9.8 m/s2 ≈ 10 m/s2) ONLY affects the y-component |
|Thus the y-component determines the time an object is in the air. |
|If gravity is the only force involved then vx,initial equals vx,final equals vx,average ( [pic] |
|[pic] [pic] |
|Example: |sin θ = oppo / hyp |cos θ = adj / hyp |2nd: Time to the top of path is same as time |
| |sin θ = vy / v |cos θ = vx / v |to the bottom of path, so |
|An USC punter kicks the ball with an initial velocity of 30 |vy = v sin θ |vx = v cos θ |ttotal = ttop + tbottom |
|m/s at a 50° angle, (so that the vy,i = 23 m/s and vx,i = 19.3|vy = 30 sin 50° |vx = 30 cos 50° | |
|m/s). What distance does the football travel? |vy = 23.0 m/s |vx = 19.3 m/s |ttotal = 2.3 + 2.3 |
| | | |ttotal = 4.6 seconds. |
|[pic] | | | |
| | | | |
| | | |vave = Δx /Δttotal |
| | | |19.3 = Δx /4.6 |
| | | |Δx = 88.8 meters |
| | | |
| |1st: Determine time to top of path. The y-comp | |
| |of the velocity at the top of the path will be 0 m/s. | |
| | | |
| |ay = Δvy / Δt | |
| |ay = (vf – vi) / Δt | |
| |10 = (0 – 23.0) / Δt | |
| |Δt = 2.3 seconds | |
| | | |
| |Note: Is this the time to the top of the path or | |
| |the time from the top to the bottom? | |
|The Range Formula |
|The range formula is a useful derivation for determining the distance traveled by objects on level ground. Below is the derivation. The above sketch is referred during the derivation. |
|ay = Δvy / Δt | ttop = tbottom |Level ground | Δx = vx-ave ( Δttotal ) |
|-g = (0 - vi) / ttop | |If the only force on the object is gravity; the |Δx = v*cosθ (2 v sinθ/g) |
|ttop = vy-ini / g |ttotal = 2ttop |time it takes to the top of the path is equal to |Δx = v2 (2cosθsinθ) / g |
| |ttotal = 2(vy-ini / g) |the time it takes from the top of the path back to|Δx = v2 (sin 2θ) / g |
| |ttotal = 2 v sinθ/g |ground level. | |
|Example |[pic] |(a) v0 = v1 = v2 > 0 |
|The figure shows the trajectory of a ball undergoing projectile | |(b) v0 = v1 > v2 = 0 |
|motion over level ground. How do the speeds v0, v1, and v2 | |(c) v0 = v1 > v2 > 0 |
|compare? | |(d) v0 > v1 > v2 > 0 |
| | |(e) v0 > v1 > v2 = 0 |
|If the figure is to scale, which component of the velocity vector is greater, vo,x and vo,y? | |What are the values of the velocity vector components, v1,x and v1,y as well as the acceleration vector |
| | |components, a1,x and a1,y and (both m/s and m/s2)? |
|What are the values of the initial acceleration vector components, ao,x and ao,y? | | |
| |[pic] |
|What is Δx? | |
|Monkey Gun, Large Demo: ME-D-ML |
| |
|Howitzer and Tunnel Demo: ME-D-HT |
| |
|Example |
| |
|A 2.00 kg ball is thrown upward with an initial speed of 20.0 m/s from the edge of a 40.0 m high cliff. At the instant the ball is thrown, a girl starts running away from the base of the cliff with a constant speed |
|of 4.00 m/s. The girl runs in a straight line on level ground. Ignore air resistance. |
|(a) |xBall = ½at2 + vo t + xo |xGirl = ½at2 + vo t + xo |
|At what angle above the horizontal should the ball be thrown so that the |xBall = 0 + cosθ 20 t + 0 |xGirl = 0 + 4 t + 0 |
|runner will catch it just before it hits the ground? | | |
| | | |
| |xBall = xGirl |As we set the x-components equal to each other, this will give us |
| |vo-ball t = vo-girl t |the angle where the x-comps of both velocities will be equal to |
| |cosθ 20 t = 4 t |each other. |
| |θ = 78.5° | |
|(b) What distance from the cliff does the girl catch the ball? |
|yBall = ½ a t2 + vo t + yo |So the ball falls to 40 meters below the cliff top 5.40 seconds after being thrown. |
|0 = ½(-10)t2 + sin78.5°20 t + 40 | |
|5t2 – 19.6 t = 40 | |
|t2 – 3.92 t = 8 | |
|(completing the squares, ½ 3.92 = 1.96) | |
|(t–1.96)2 = 8 + (1.96)2 | |
|t – 1.96 = ±3.44 | |
|t = 1.96 ± 3.44 | |
|t = 5.40 seconds | |
| |xGirl = 3.47 t |
| |xGirl = 3.47 m/s (6.82sec) |
| |xGirl = 23.7 meters |
|4.4 Uniform Circular Motion |
|With this as a base let’s follow a |[pic] |
|few steps: | |
|Draw in two radii vectors, r1 & r2 | |
|at an angular displacement of θ |[pic] |
|Draw in change of position of the | |
|radii vectors, Δr | |
|At the position of radii vectors r1|[pic] |
|& r2, let draw in velocity vectors | |
|v1 & v2 | |
|Next, displace so that velocity | |
|vector tails so that they are | |
|adjacent to each other. | |
|Since vector v1 is perpendicular to| |
|r1 & v2 is perpendicular to r2 the | |
|angle, θ, between r1 & r2 must be | |
|the same as the angle between v1 & | |
|v2 | |
| | |
|Since the two triangles are | |
|“similar” and follow all the rules | |
|of similar triangles and |v1| = | |
||v2| (so we can write v1 or v2 as | |
|simply v) we write the following | |
|equation | |
|[pic] or [pic] which can be | |
|rewritten as [pic] | |
| |
|At this point let’s substitute back into a = Δv /Δt |
|[pic] ; [pic]; but we know [pic]; so that [pic] ( ac = v2 / r |
|Question: |Ac = v2 / r |
|If a vehicle is in a turn |ac = 20.02 / 30.0 |
|of radius 30.0 meters and |ac = 13.3 m/s2 |
|is traveling at 20.0 m/s | |
|what is the centripetal | |
|acceleration required for | |
|this car to remain in this| |
|turn? | |
|At this we need to |An alternate method to write centripetal |
|introduce the period T, |acceleration is |
|the amount of time for one|[pic] |
|complete revolution (or | |
|iteration, process, etc). | |
|If an object is in a | |
|circular path the amount | |
|of time to complete path | |
|of one revolution is the | |
|period and the distance to| |
|complete the same | |
|revolution is 2πr. | |
|[pic]; where (r = 2πr and | |
|(t = T ( [pic] | |
|Results of this discussion: |
|An object with constant speed (remember speed is magnitude only) while |
|traveling in a circle has a constant centripetal acceleration. |
|Does this mean you slow down much more rapidly in a circular path in |
|your vehicle? |
|Try it some time…release your gas pedal on a flat surface and note your|
|rate of “deceleration” (negative acceleration) |
|Then compare this to your rate of negative acceleration when you are in|
|a turn. |
|Ans: You will notice your car slows down much more rapidly in a turn |
|due to centripetal acceleration |
|So to maintain your constant speed in a circle you must press down more|
|so (or positively accelerate) on your accelerator to maintain a |
|constant speed shown by your speedometer |
|Demo: Twirl-a-Bob: ME-D-TB |
|4.5 Tangential and Radial Acceleration |
| | |
|Key |[pic] |
|Point to|The radial acceleration component arises from the change in |
|this |direction of the velocity vectors as shown above. |
|section | |
|Radial |Since centripetal means “Towards” the center (as opposed to |
|Unit |centrifugal which is “Away” from the center) we know the |
|Vector, |radial component of acceleration is the centripetal |
|[pic] |acceleration. |
|always | |
|points |Ar = -ac = -v2/r |
|away |The negative “-“sign represents the opposing direction of the |
|from the|centripetal acceleration (toward the center) to the radial |
|center |unit vector, [pic]. |
|(as | |
|shown | |
|above) | |
|Results | |
|of this | |
|discussi| |
|on | |
|ac is | |
|toward | |
|the | |
|center | |
|ar is | |
|toward | |
|the | |
|center | |
|is away | |
|from the| |
|center | |
|Think |Tangential Acceleration: As the name indicates tangential |
|About It|acceleration is an acceleration tangent to the radius. |
| |[pic] |
|If you |If an object has a tangential acceleration component vector, |
|release |then the tangential velocity will increase with respect to |
|(remove |time. This will result in an object “speeding up/slowing |
|the |down”. |
|centripe| |
|tal | |
|force) | |
|an | |
|object | |
|that | |
|travelin| |
|g in a | |
|circular| |
|path, | |
|the | |
|object | |
|goes off| |
|in a | |
|straight| |
|line | |
|which is| |
|tangent | |
|to the | |
|circle | |
|at that | |
|point. | |
|(Demonst| |
|ration | |
|in | |
|lecture)| |
|Results of this discussion |
|1. If the object|2. If the object has a constant centripetal |
|is in a circular |acceleration and a ZERO tangential acceleration, the |
|path and has a |object maintains a constant speed. |
|tangential | |
|acceleration, | |
|this means the | |
|object does NOT | |
|have a constant | |
|speed. | |
|Equ| [pic] |Example |Ans: | |
|ati| |What is the |a2 = |[pic] |
|on | |acceleration of |ar2 + | |
| | |the object in the|aT2 | |
|a2 | |picture the left |a2 = 82| |
|= | |if aT = 3.0 m/s2 |+ 32 | |
|ar2| |and ar = 8.0 ||a| = | |
|+ | |m/s2? |8.5 m/s2| |
|aT2| | |@ | |
| | | |tan ( = | |
| | | |3 / 8 | |
| | | |( = | |
| | | |20.6( | |
| | | |off | |
| | | |center | |
|[pic] |
|4.6 Relative Velocity and Acceleration |
|Reference frames!!! |To the student walking it appears to go straight|
|Let’s pretend you are|up and down. |
|back in 7th grade. | |
|You are on the bus |Which is true? |
|coming to your | |
|favorite |Both, depending on your reference frame. |
|place…SCHOOL!!! | |
| | |
|As you get close to |r’ = r + vot |
|school you pass the | |
|kids that are walking|r is the reference frame of the walker |
|to school. One of |r’ is the reference from of the bus with vo |
|these kids loves to |busing the velocity of the bus. |
|throw his baseball up| |
|and down will | |
|walking. | |
| | |
|What do you see? | |
| | |
|You see a ball that | |
|rises as you approach| |
|it…and you pass the | |
|ball…it recedes in | |
|the distance on it’s | |
|way down. | |
| | |
|The ball follows a | |
|parabolic path to | |
|you. | |
|Copied from 3.7 Relative Motion |
|15 + 1.2 = 16.2 |[pic] |
|m/s | |
| | |
|It appears the | |
|hobo in the train| |
|car is moving | |
|+16.2 m/s from | |
|the train | |
|engineer’s | |
|perspective | |
|15 – 1.2 = 13.8 | |
|m/s |[pic] |
| | |
|It appears the | |
|hobo in the train| |
|car is moving | |
|+13.8 m/s from | |
|the train | |
|engineer’s | |
|perspective | |
|[pic] |[pic] |
|a2 + b2 = c2| |
| | |
|or | |
| | |
|vtg2 + vpt2 | |
|= vpg2 | |
|Example |
|A ball of mass 0.2 kg has a velocity of 1.50 i m/s; a ball of mass 0.3 kg has a velocity of -0.4 i m/s. They meet in a head-on collision. (a) Find their velocities after the |
|collision. (b) Find the velocity of their center of mass before and after the collision. |
| |
|Use| |v1 - 1.5 = 0 |
|rel| |v2 + 0.4 = 0 (subtract the two equations) |
|ati| |v1 - v2 -1.9 = 0 |
|ve | | or v1 = 1.9 + v2 |
|vel| |This is before collision |
|oci| |After the collision, it is reversed |
|tie| | v2f = 1.9 + v1f |
|s | | |
|fro| | |
|m | | |
|ch | | |
|4 | | |
| | | |
|If | | |
|v1 | | |
|= | | |
|1.5| | |
|and| | |
|v2 | | |
|= | | |
|-0.| | |
|4 | | |
|the| | |
|n | | |
|It | | |
|app| | |
|ear| | |
|s | | |
|to | | |
|v1 | | |
|tha| | |
|t | | |
|v2 | | |
|is | | |
|app| | |
|roa| | |
|chi| | |
|ng | | |
|at | | |
|1.9| | |
|m/s| | |
|as | | |
|if | | |
|v1 | | |
|is | | |
|not| | |
|mov| | |
|ing| | |
|. | | |
| | | |
|m1 v1 + m2 v2 = m1f v1f + m2f v2f |(b) |
|0.2*1.5 + 0.3*(-.4) = 0.2 * v1f + 0.3 v2f |vCM = (m1fv1f + m2fv2f) / m |
| |vCM = 0.2*1.5 + 0.3*(-4)/5 |
|Use relative velocities from section 4.6 |vCM = +0.36 m/s |
| v1-init = 1.9 + v2-init then v2f = 1.9 m/s + v1f | |
| |This is vCM before the collision…and since momentum is conserved, this is |
|0.2*1.5 + 0.3*(-.4) = 0.2 * v1f + 0.3(1.9 m/s + v1f) |also the vCM after the collision. |
|0.3 - 0.12 = 0.2v1f + 0.57 + 0.3v1f | |
|0.3 - 0.12 = 0.5v1f + 0.57 | |
|v1f = -0.78 m/s | |
|v2f = 1.12 m/s | |
| | |
|This gives you a hint of the fun we will be encountering in Chapter 9 | |
[pic]
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