Monday, June 14, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu1PHYS 1441 – Section 501Lecture #4Monday, June 14, 2004Dr. Jaehoon YuFirst term exam at 6pm, next Wednesday, June 23!!• Chapter three: Motion in two dimension– Two dimensional equation of motion– Projectile motion• Chapter four: Newton’s Laws of Motion– Force and Mass– Newton’s Laws of Motion– Solving problems using Newton’s Laws– Friction ForcesMonday, June 14, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu2Announcements• E-mail distribution list: 36 of you have registered– This Wednesday is the last day of e-mail registration– -5 extra points if you don’t register by next Wednesday, June 23– A test message was sent out today.• Term exam:– Date: Next Wednesday, June 26–Time: In the class, 6 – 7:50pm–Covers: Chapters 1 – 6.5 and Appendix A– Location: Classroom, SH 125– Mixture of multiple choices and numeric calculations– There will be a total of 3 exams of which two best will be chose for your grades– Missing an exam is not permitted unless approved prior to the exam by me Î I will only be in e-mail contact for this exam.• Class Schedule:– Dr. White will teach you on June 16 and 21– Mr. Kaushik (Rm 133, x25700) will proctor your exam on June 23Monday, June 14, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu3Instantaneous AccelerationAverage AccelerationInstantaneous VelocityAverage VelocityDisplacementTwo DimensionsOne DimensionQuantityDisplacement, Velocity, and Acceleration in 2-dimifrrr −=∆ififttrrtrv−−=∆∆≡0limtrvt∆→∆≡∆GGififttvvtva−−=∆∆≡0limtvat∆→∆≡∆GGixxx f−≡∆txttxxviixff∆∆=−−≡xxvt→∆=∆Δt0limxfxfxixivv vatt t−∆≡=−∆xxvat→∆≡∆Δt0limMonday, June 14, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu42-dim Motion Under Constant Acceleration• Position vectors in x-y plane:ir=Gfr=G• Velocity vectors in x-y plane:iv=Gfv=Gxfv=• How are the position vectors written in acceleration vectors?fr =Gfx=fv=Gyfv=fy=()iixi y j=+GGVelocity vectors in terms of acceleration vectoriixiyj+GGffxi y j+GGxi yivi v j+GGxf yfvi v j+GGxixvat+yi yvat+()()xi x yi yvativatj+++ =GGivat+GG212ixi xxvt at++212iyi yyvt at++fxiGfyj+=G212ixi xxvt ati⎛⎞++⎜⎟⎝⎠G212iyi yyvt atj⎛⎞+++⎜⎟⎝⎠G()xi yivi v j t++GG()212xyai a j t++GGir=JGivt+G212at+GMonday, June 14, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu5Example for 2-D Kinematic EquationsA particle starts at origin when t=0 with an initial velocity vv=(20ii-15jj)m/s. The particle moves in the xy plane with ax=4.0m/s2. Determine the components of velocity vector at any time, t.xfvCompute the velocity and speed of the particle at t=5.0 s.5tv=G()()22xyspeed v v v== +G()vtGyfvxiv=xat+20=()4.0 /tms+yiv=yat+15=− 0t+()15 /ms=−Velocity vector()xvti=G()yvtj+G()20 4.0ti=+G15 ( / )jm s−G,5xtvi==G,5ytvj=+G()20 4.0 5.0 i=+×G15j−G()40 15 /ijms=−GG() ( )2240 15 43 /ms=+−=Monday, June 14, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu6Example for 2-D Kinematic Eq. Cnt’dθ=Determine the x and y components of the particle at t=5.0 s.fx =fr =Gfy =Can you write down the position vector at t=5.0s?Angle of the Velocity vector1tanyxvv−⎛⎞=⎜⎟⎝⎠115tan40−−⎛⎞=⎜⎟⎝⎠13tan 218−−⎛⎞=−⎜⎟⎝⎠Dxivt212xat+20 5=×21452+×× =150( )myivt=15 5−×=75 ( )m−fxiGfyj+G150i=G()75 jm−GMonday, June 14, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu7Projectile Motion The only acceleration in this motion. It is a constant!!Monday, June 14, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu8Projectile Motion• A 2-dim motion of an object under the gravitational acceleration with the assumptions– Gravitational acceleration, -g, is constant over the range of the motion– Air resistance and other effects are negligible• A motion under constant acceleration!!!! Î Superposition of two motions– Horizontal motion with constant velocity and– Vertical motion under constant accelerationa =Gt=Show that a projectile motion is a parabola!!!xiv =fx=fy=2cos21cossin⎟⎟⎠⎞⎜⎜⎝⎛−⎟⎟⎠⎞⎜⎜⎝⎛=iifiifiifvxgvxvyθθθPlug in the t aboveIn a projectile motion, the only acceleration is gravitational one whose direction is always toward the center of the earth (downward).ax=0What kind of parabola is this?222cos2tanfiiiffxvgxy⎟⎟⎠⎞⎜⎜⎝⎛−=θθcosivιθyiv=siniivθxyai aj+=GGgj−Gxivt= cosiivtθcosfiixvθ=21sin2iivtgtθ=−()21 2yivtgt+−Monday, June 14, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu9Example for Projectile MotionA ball is thrown with an initial velocity vv=(20ii+40jj)m/s. Estimate the time of flight and the distance the ball is from the original position when landed.Which component determines the flight time and the distance?fy=fx =Flight time is determined by y component, because the ball stops moving when it is on the ground after the flight.Distance is determined by xcomponent in 2-dim, because the ball is at y=0 position when it completed it’s flight.sec880or 0 ≈==∴gtt()80 0tgt−=()21402tgt+−=0m8sect∴≈xivt=()20 8 160 m×=Monday, June 14, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu10Horizontal Range and Max Height• Based on what we have learned in the previous pages, one can analyze a projectile motion in more detail– Maximum height an object can reach– Maximum rangeyfv=At the maximum height the object’s vertical motion stops to turn around!!fy=Since no acceleration in x, it still flies even if vy=02xi ARvt==vviiθhgvtiAιθsin=∴⎟⎟⎠⎞⎜⎜⎝⎛=gvRiiθ2sin22sin21sinsin⎟⎟⎠⎞⎜⎜⎝⎛−⎟⎟⎠⎞⎜⎜⎝⎛=gvggvvyiiiiiifθθθ⎟⎟⎠⎞⎜⎜⎝⎛=gvyiif2sin22θWhat happens at the maximum height?sin2cosiiiivvgθθ⎛⎞⎜⎟⎝⎠yi yvat+=siniAvgtιθ−=0h =()212yivt gt=+−Monday, June 14, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu11Maximum Range and Height• What are the conditions that give maximum height and range of a projectile motion?⎟⎟⎠⎞⎜⎜⎝⎛=gvhii2sin22θThis formula tells us that the maximum height can be achieved when θi=90o!!!⎟⎟⎠⎞⎜⎜⎝⎛=gvRiiθ2sin2This formula tells us that the maximum range can be achieved when 2θi=90o, i.e., θi=45o!!!Monday, June 14, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu12Example for Projectile Motion• A stone was thrown upward from the top of a building at an angle of 30oto horizontal with initial speed of 20.0m/s. If the height of the building is
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