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9/9 2-D Motion, Projectile Motion  Text sections and 1.6-8, 3.1-2, and 4.3, 4.7 Ch. 4 is about forces and gravity  HW “9/9 2-D Motion” Due Thursday,

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Presentation on theme: "9/9 2-D Motion, Projectile Motion  Text sections and 1.6-8, 3.1-2, and 4.3, 4.7 Ch. 4 is about forces and gravity  HW “9/9 2-D Motion” Due Thursday,"— Presentation transcript:

1 9/9 2-D Motion, Projectile Motion  Text sections and 1.6-8, 3.1-2, and 4.3, 4.7 Ch. 4 is about forces and gravity  HW “9/9 2-D Motion” Due Thursday, 9/12 On web or in 213 Witmer for copying  Lab: “Vector Exercise” watch notation in lab  Suggested problems: 3-25, 27, 30 Use concepts, not equations!!!

2 Describing 2-D Motion A bead moves right on a wire at 3m/s and the wire moves down at 4m/s. Find the actual velocity of the bead, magnitude and direction. v y = 4m/s v x = 3m/s v = 5m/s  = 53° below horizontal The acceleration in the x direction is 4m/s 2 left and the acceleration in the y direction is 0m/s 2. Find the actual velocity of the bead 2 seconds later. Specify both magnitude and direction.

3 Describing 2-D Motion v y = 4m/s a y = 0m/s 2 v x = 3m/s a x = 4m/s 2 left v = 5m/s at t = 0s  = 53° below horizontal at t = 0s At t = 2s: v y = 4m/s down a y = 0m/s 2 v x = 5m/s left a x = 4m/s 2 left v =  4 2 + 5 2 = 6.4m/s  = Arctan 4/5 = 39° below horizontal

4 Example: v y = ? v x = ? v = 20m/s  = 53° above horizontal  An object has an initial velocity of 20m/s directed at 53  above the horizontal. The acceleration is 4m/s 2 directed down. What is the object’s velocity 8s later? 12m/s right 16m/s up a x = 0 so v x at 8s = 12m/s right a y = 4m/s 2 down so v y at 8s = 16m/s down v x = 12m/s right v = 20m/s  = 53° below horizontal  v y = 16m/s down At what time is it moving exactly horizontal? What is its speed at this time? 4s 12m/s

5 Example: another look v y = ? v x = ? v = 20m/s  = 53° above horizontal  An object has an initial velocity of 20m/s directed at 53  above the horizontal. The acceleration is 4m/s 2 directed down. What is the object’s velocity 8s later? 12m/s right 16m/s up v x = 12m/s right v = 20m/s  = 53° below horizontal  v y = 16m/s down Just like we do for 1D motion vivi vv vfvf v i +  v = v f

6 g is not the “accel. of gravity” g is always a “field strength”, not an acceleration! It tells us how much force (in Newtons) is on an object and depends on the object’s mass. Weight = mass x g An object in free fall (or in projectile motion) has only the weight force on it as nothing else is touching it.

7 g is not the “accel. of gravity” Newton’s 2nd Law It tells us how much acceleration an object has and depends on the object’s mass as well as the force on the object. acceleration = force (Weight)  mass Be careful: g = 9.8N/kg and for free fall a = 9.8m/s 2 but it is dangerous to think of g as an acceleration. More on this later. a = W/m = mg/m = ga = g in number, not in concept!

8 Projectile Motion Example: An object is thrown with a velocity of 4m/s up and 3m/s west. The acceleration is 9.8m/s 2 directed down. How long is the object in the air? How high does it get? What is its velocity when it gets back to the ground?


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