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## PHYS 1441 – Section 001 Lecture #6

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**PHYS 1441 – Section 001Lecture #6**Wednesday, June 17, 2015 Dr. Jaehoon Yu • Understanding the 2 Dimensional Motion • 2D Kinematic Equations of Motion • Projectile Motion • Maximum Range and Height PHYS 1441-001, Summer 2015 Dr. Jaehoon Yu**Announcements**• Quiz #2 • Beginning of the class tomorrow, Thursday, June 18 • Covers CH3.1 to what we finish today, Wednesday, June 17 • Bring your calculator but DO NOT input formula into it! • You can prepare a one 8.5x11.5 sheet (front and back) of handwrittenformulae and values of constants for the exam no solutions of any kind, derivations or word definitions! • No additional formulae or values of constants will be provided! PHYS 1441-001, Summer 2015 Dr. Jaehoon Yu**Reminder: Special Project#2 for Extra Credit**• Show that the trajectory of a projectile motion is a parabolaand explain your answer. • 20 points • Due: Monday,June 22 • You MUST show full details of your OWN computations to obtain any credit • Beyond what was covered in this lecture noteand in the book! PHYS 1441-001, Summer 2015 Dr. Jaehoon Yu**2D Displacement**PHYS 1441-001, Summer 2015 Dr. Jaehoon Yu**2D Average Velocity**Average velocity is the displacement divided by the elapsed time. What is the direction of the Average velocity? Right! The same as the displacement! PHYS 1441-001, Summer 2015 Dr. Jaehoon Yu**The instantaneous velocity indicates how fast the car moves**and the direction of motion at each instant of time. 2D Instantaneous Velocity PHYS 1441-001, Summer 2015 Dr. Jaehoon Yu**2D Average Acceleration**PHYS 1441-001, Summer 2015 Dr. Jaehoon Yu**Displacement, Velocity, and Acceleration in 2-dim**• Average Velocity: • Displacement: How is each of these quantities defined in 1-D? • Instantaneous Velocity: • Average Acceleration • Instantaneous Acceleration: PHYS 1441-001, Summer 2015 Dr. Jaehoon Yu**Kinematic Quantities in 1D and 2D**What is the difference between 1D and 2D quantities? PHYS 1441-001, Summer 2015 Dr. Jaehoon Yu**A Motion in 2 Dimension**This is a motion that could be viewed as two motions (x and y directions) combined into one. (superposition…) PHYS 1441-001, Summer 2015 Dr. Jaehoon Yu**Motion in horizontal direction (x)**PHYS 1441-001, Summer 2015 Dr. Jaehoon Yu**Motion in vertical direction (y)**PHYS 1441-001, Summer 2015 Dr. Jaehoon Yu**A Motion in 2 Dimension**Imagine you add the two 1 dimensional motions on the left. It would make up a one 2 dimensional motion on the right. PHYS 1441-001, Summer 2015 Dr. Jaehoon Yu**Kinematic Equations in 2-Dim**y-component x-component PHYS 1441-001, Summer 2015 Dr. Jaehoon Yu**Ex. A Moving Spacecraft**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. PHYS 1441-001, Summer 2015 Dr. Jaehoon Yu**How do we solve this problem?**• Visualize the problem Draw a picture! • Decide which directions are to be called positive (+) and negative (-). Normal convention is the right-hand rule. • Write down the values that are given for any of the five kinematic variables associated with each direction. • 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. • When the motion is divided into segments, remember that the final velocity of one segment is the initial velocity for the next. • Keep in mind that there may be two possible answers to a kinematics problem. PHYS 1441-001, Summer 2015 Dr. Jaehoon Yu**Ex. continued**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. +24.0 m/s2 +22.0 m/s 7.0 s +12.0 m/s2 +14.0 m/s 7.0 s PHYS 1441-001, Summer 2015 Dr. Jaehoon Yu**First, the motion in x-direciton…**PHYS 1441-001, Summer 2015 Dr. Jaehoon Yu**Now, the motion in y-direction…**PHYS 1441-001, Summer 2015 Dr. Jaehoon Yu**The final velocity…**A vector can be fully described when the magnitude and the direction are given. Any other way to describe it? Yes, you are right! Using components and unit vectors!! PHYS 1441-001, Summer 2015 Dr. Jaehoon Yu**If we visualize the motion…**PHYS 1441-001, Summer 2015 Dr. Jaehoon Yu**What is the 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 • and • Air resistance and other effects are negligible • A motion under constant acceleration!!!! Superposition of two motions • Horizontal motion with constant velocity ( no acceleration ) • Vertical motion under constant acceleration ( g ) PHYS 1441-001, Summer 2015 Dr. Jaehoon Yu**The only acceleration in this motion. It is a constant!!**Projectile Motion But the object is still moving!! Why? Because it still has constant velocity in horizontal direction!! Why? Because there is no acceleration in horizontal direction!! Maximum height PHYS 1441-001, Summer 2015 Dr. Jaehoon Yu**Kinematic Equations for a projectile motion**y-component x-component PHYS 1441-001, Summer 2015 Dr. Jaehoon Yu**Example for a Projectile Motion**A ball is thrown with an initial velocity v=(20i+40j)m/s. Estimate the time of flight and the distance from the original position when the ball lands. Which component determines the flight time and the distance? Flight time is determined by the y component, because the ball stops moving when it is on the ground after the flight. So the possible solutions are… Distance is determined by the x component in 2-dim, because the ball is at y=0 position when it completed it’s flight. Why isn’t 0 the solution? PHYS 1441-001, Summer 2015 Dr. Jaehoon Yu**Ex.3.6 The Height of a Kickoff**A placekicker kicks a football at an angle of 40.0 degrees and the initial speed of the ball is 22 m/s. Ignoring air resistance, determine the maximum height that the ball attains. PHYS 1441-001, Summer 2015 Dr. Jaehoon Yu**First, the initial velocity components**PHYS 1441-001, Summer 2015 Dr. Jaehoon Yu**Motion in y-direction is of the interest..**? -9.8 m/s2 0 m/s +14 m/s PHYS 1441-001, Summer 2015 Dr. Jaehoon Yu**Now the nitty, gritty calculations…**What happens at the maximum height? The ball’s velocity in y-direction becomes 0!! Because there is no acceleration in x-direction!! And the ball’s velocity in x-direction? Stays the same!! Why? Which kinematic formula would you like to use? Solve for y PHYS 1441-001, Summer 2015 Dr. Jaehoon Yu**Ex.3.6 extended: The Time of Flight of a Kickoff**What is the time of flight between kickoff and landing? PHYS 1441-001, Summer 2015 Dr. Jaehoon Yu**What is y when it reached the max range?**0 m -9.80 m/s2 14 m/s PHYS 1441-001, Summer 2015 Dr. Jaehoon Yu**Now solve the kinematic equations in y direction!!**Since y=0 or Two soultions Solve for t PHYS 1441-001, Summer 2015 Dr. Jaehoon Yu**Ex.3.9 The Range of a Kickoff**Calculate the range R of the projectile. PHYS 1441-001, Summer 2015 Dr. Jaehoon Yu**Example for a Projectile Motion**• A stone was thrown upward from the top of a cliff at an angle of 37o to horizontal with initial speed of 65.0m/s. If the height of the cliff is 125.0m, how long is it before the stone hits the ground? Becomes Since negative time does not exist. PHYS 1441-001, Summer 2015 Dr. Jaehoon Yu 34**Example cont’d**• What is the speed of the stone just before it hits the ground? • What are the maximum height and the maximum range of the stone? Do these yourselves at home for fun!!! PHYS 1441-001, Summer 2015 Dr. Jaehoon Yu 35