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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