Thursday, June 9, 2011 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 1 PHYS 1443 – Section 001 Lecture #4 Thursday, June 9, 2011 Dr. Jaehoon Yu • Motion in Two Dimensions: • Motion under a constant acceleration • Projectile Motion • Maximum range and Maximum height • Newton’s Laws of Motion • Force; Mass; Newton’s 1st – 3rd LawsThursday, June 9, 2011 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 2 Announcements • Homework – 100% of you have registered and submitted!! – Excellent job! – 6/27 already started working on HW#2, Great!! • Quiz 1 results – Class average: 7.5/12 • Equivalent to 62.5/100 – Top score: 11.2/12 • Quiz #2 coming Tuesday, June 14 – Covers: CH 1.1 – what we finish on Monday, June 13Thursday, June 9, 2011 3 Reminder: Special Project for Extra Credit • Show that the trajectory of a projectile motion is a parabola!! – 20 points – Due: next Tuesday, June 14 – You MUST show full details of your OWN computations to obtain any credit • Much beyond what was covered in page 21 of this lecture note!! PHYS 1443-001, Spring 2011 Dr. Jaehoon YuThursday, June 9, 2011 This is a motion that could be viewed as two motions combined into one. (superposition…) A Motion in 2 Dimension PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 4Thursday, June 9, 2011 Motion in horizontal direction (x) x =12vxo+ vx( )t x = vxot +12axt2 vx= vxo+ axt vx2= vxo2+ 2axxPHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 5Thursday, June 9, 2011 Motion in vertical direction (y) y =12vyo+ vy( )t y = vyot +12ayt2 vy= vyo+ ayt vy2= vyo2+ 2ayyPHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 6Thursday, June 9, 2011 Imagine you add the two 1 dimensional motions on the left. It would make up a one 2 dimensional motion on the right. A Motion in 2 Dimension PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 7Thursday, June 9, 2011 x-component x =12vxo+ vx( )t Δx = vxot +12axt2 vx= vxo+ axt vx2= vxo2+ 2axx y =12vyo+ vy( )t Δy = vyot +12ayt2 vy= vyo+ ayt vy2= vyo2+ 2ayyKinematic Equations in 2-Dim y-component PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 8Thursday, June 9, 2011 In the x direction, the spacecraft in zero-gravity zone has an initial velocity component of +22 m/s and an acceleration of +24 m/s2. In the y direction, the analogous quantities are +14 m/s and an acceleration of +12 m/s2. Find (a) x and vx, (b) y and vy, and (c) the final velocity of the spacecraft at time 7.0 s. Ex. A Moving Spacecraft PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 9Thursday, June 9, 2011 How do we solve this problem? 1. Visualize the problem Draw a picture! 2. Decide which directions are to be called positive (+) and negative (-). 3. Write down the values that are given for any of the five kinematic variables associated with each direction. 4. Verify that the information contains values for at least three of the kinematic variables. Do this for x and y separately. Select the appropriate equation. 5. When the motion is divided into segments in time, remember that the final velocity of one time segment is the initial velocity for the next. 6. Keep in mind that there may be two possible answers to a kinematics problem. PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 10Thursday, June 9, 2011 In the x direction, the spacecraft in a zero gravity zone has an initial velocity component of +22 m/s and an acceleration of +24 m/s2. In the y direction, the analogous quantities are +14 m/s and an acceleration of +12 m/s2. Find (a) x and vx, (b) y and vy, and (c) the final velocity of the spacecraft at time 7.0 s. x ax vx vox t ? ? y ay vy voy t ? ? Ex. continued +24.0 m/s2 +22.0 m/s +14.0 m/s +12.0 m/s2 7.0 s 7.0 s PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 11Thursday, June 9, 2011 x ax vx vox t ? +24.0 m/s2 ? +22 m/s 7.0 s Δx = vx= voxt +12axt2 = 22 m s( )7.0 s( )+1224 m s2( )7.0 s( )2= +740 m vox+ axt = 22 m s( )+ 24 m s2( )7.0 s( )= +190 m sFirst, the motion in x-direciton… PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 12Thursday, June 9, 2011 y ay vy voy t ? +12.0 m/s2 ? +14 m/s 7.0 s Now, the motion in y-direction… Δy = voyt +12ayt2 = 14 m s( )7.0 s( )+1212 m s2( )7.0 s( )2= +390 m voy+ ayt = 14 m s( )+ 12 m s2( )7.0 s( )= +98m sPHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 13Thursday, June 9, 2011 v vy= 98m s vx= 190 m sθθ= 190 m s( )2+ 98 m s( )2= 210 m sThe final velocity… tan−198 190( )= 27A vector can be fully described when the magnitude and the direction are given. Any other way to describe it? Yes, you are right! Using the components and unit vectors!! vxi + vyj = 190i + 98j( ) m sPHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 14Thursday, June 9, 2011 If we visualize the motion… PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 15Thursday, June 9, 2011 2-dim Motion Under Constant Acceleration • Position vectors in x-y plane: • Velocity vectors in x-y plane: Velocity vectors in terms of the acceleration vector vxi+ axt( )i + vyi+ ayt( )j = =vi+atX-comp Y-comp vxii + vyij( )+ axi + ayj( )t =PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 16Thursday, June 9, 2011 2-dim Motion Under Constant Acceleration • How are the 2D position vectors written in acceleration vectors? Position vector components Putting them together in a vector form Regrouping the above = xii + yij( )2D problems can be interpreted as two 1D problems in x and y PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 17Thursday, June 9, 2011 Example for 2-D Kinematic Equations Compute the velocity and the speed of the particle at t=5.0 s. vt =5 speed = v= vx( )2+ vy( )2Velocity vector = vx ,t =5i +vy,t =5j = 20 + 4.0 × 5.0( )i −15j = 40i − 15j( ) m / sPHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 18Thursday, June 9, 2011 Example for 2-D Kinematic Eq. Cnt’d Determine the x and y components of the particle at t=5.0 s. rf=Can you write down the position vector at t=5.0s? Angle of the Velocity vector xfi + yfj = 150i −75 jm( )PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 19Thursday, June 9, 2011 Projectile Motion • A 2-dim motion of an object under the gravitational acceleration with the following assumptions – Free fall acceleration, g, is constant over the range of the motion • – Air resistance and other effects are negligible • A motion under constant acceleration!!!!
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