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Slide 2 Motion in Two Dimensions: Projectile Motion Circular Motion Angular Speed Simple Harmonic Motion Torque Center of Mass Slide 3 Projectile Motion A red marble is dropped off a cliff at the same time a black one is shot horizontally. At any point in time the marbles are at the same height, i.e., theyre falling down at the same rate, and they hit the ground at the same time. Gravity doesnt care that the black ball is moving sideways; it pulls it downward just the same. Since gravity cant affect horiz. motion, the black particle continues at a constant rate. With every unit of time, the marbles vertical speed increases, but their horiz. speed remains the same (ignoring air resistance). continued on next slide Slide 4 Projectile Motion continued on next slide Gravitys downward pull is independent of horiz. motion. So, the vertical acceleration of each marble is - g (for the whole trip), and the sideways acceleration of each is zero. (Gravity cant pull sideways). Whatever horiz. velocity the black one had when shot is a constant throughout its trip. Only its vertical velocity changes. (A vertical force like gravity can only produce vertical acceleration.) 9.8 m/s 2 Slide 5 Projectile Motion (cont.) continued on next slide t = 0 t = 1 t = 2 t = 3 t = 4 v y = 1 v y = 2 v y = 3 v y = 4 v y = 0 If after one unit of time the marbles have one unit of speed downward, then after two units of time they have two units of speed downward, etc. This follows directly from v f = v 0 + a t. Since v 0 = 0, downward speed is proportional to time. Note: The vectors shown are vertical components of velocity. The shot marble has a horizontal component too (not shown); the dropped one doesnt. Slide 6 Since the shot black marble experiences no horiz. forces (ignoring air), it undergoes no horiz. acceleration. Therefore, its horiz. velocity, doesnt change. So, the horiz. vector has a constant magnitude, but the vertical vector gets longer. The resultant (the net velocity vector in blue) gets longer and points more downward with time. When t = 0, v = v x for the shot marble. v = v y for the dropped marble for the whole trip. Projectile Motion (cont.) continued on next slide v x = v vxvx vxvx vxvx vxvx v v v v y t = 0 t = 1 t = 2 t = 3 t = 4 Slide 7 Projectile Motion (cont.) The trajectory of any projectile is parabolic. (Well prove this later.) If its initial velocity vector is horizontal, as with the black marble, the launch site is at the vertex of the parabola. The velocity vector at any point in time is tangent to the parabolic trajectory. Moreover, velocity vectors are always tangent to the trajectory of any moving object, regardless of its shape. v v v continued on next slide Slide 8 x = 1 Projectile Motion (cont.) y = 1 y = 3 y = 5 y = 7 x = 2 x = 3 x = 4 continued on next slide The vertical displacements over consecutive units of time show the familiar ratio of odd numbers that weve seen before with uniform acceleration. Measured from the starting point, the vertical displacements would be 1, 4, 9, 16, etc., (perfect squares), but the horiz. displacements form a linear sequence since there is no acceleration in that direction. t = 0 t = 1 t = 2 t = 3 t = 4 Slide 9 Projectile Example A rifle is held perfectly horizontally 1.5 m over level ground. At the instant the trigger is pulled, a second bullet is dropped from the tip of the barrel. The muzzle velocity of the gun is 80 m/s. 1. Which bullet hits the ground first? answer: 2. How fast is each bullet moving after 0.3 s ? answer: dropped bullet after 0.3 s fired bullet after 0.3 s 80 m/s They hit at same time. vyvy vyvy Use v f = v 0 + a t and use vertical info only: v 0 = 0, a = -9.8 m/s 2, and t = 0.3 s. We get v y in the pic for each bullet is -2.94 m/s. Using the Pythagorean theorem for the fired bullet we get 80.054 m/s in a direction tangent to its path. continued on next slide Slide 10 Projectile Example (cont.) 80 m/s 1.5 m 3. How far away does the fired bullet land (its range)? answer: The first step is to find the its hang time. This is the same hang time as the dropped bullet. Use y = v 0 t + 0.5 a t 2 with only vertical data: -1.5 = (0) t + (0.5) (-9.8) t 2. So, t = 0.5533 s. The whole time the bullet is falling its also moving to the left at a constant 80 m/s. Since horizontally v is constant, we use d = v t with only horiz. info: d = (80 m/s) (0.5533 s) = 44.26 m. Note: When a = 0, x = v 0 t + 0.5 a t 2 breaks down to d = v t. Slide 11 Now lets find range of a projectile fired with speed v 0 at an angle . Step 1: Split the initial velocity vector into components. Projectiles Fired at an Angle v 0 cos v 0 sin v0v0 continued on next slide Slide 12 Step 2: Find hang time. Use y = v 0 t + a t 2 with only vertical data: v 0 cos v 0 sin v0v0 y = ( v 0 sin ) t + (-g)t 2 Over level ground, y = 0. Divide through by t : 0 = v 0 sin - 4.9 t, and t = ( v 0 sin ) / 4.9 Note: If we had shot the projectile from a 100 m cliff, y would be -100 m. continued on next slide Projectiles Fired at an Angle (cont.) Slide 13 v 0 cos Step 3: Now that we know how long its in the air, we know how long it travels horizontally. (The projectiles vertical and horizontal movements are completely independent.) Use x = v 0 t + a t 2 again, this time with only horizontal data: v 0 sin v0v0 x = (v 0 cos ) t + (0) t 2 = (v 0 cos ) t This is the same as saying: horiz. distance = horiz. speed time In other words, d = v t continued on next slide Projectiles Fired at an Angle (cont.) Slide 14 Picklemobile Example A stuntman drives a picklemobile off a 350 m cliff going 70 mph. The angle of elevation of the cliff is 21 . Hes hoping to make it across a 261 m wide river and land on a ledge 82 m high. Does he make it ? 70 mph 350 m 82 m Well, the first thing we have to do is convert the initial velocity into m/s: 70 mi h 1609 m mi h 3600 s = 31.2861 m/s 261 m continued on next slide 21 Slide 15 Picklemobile Example (cont.) 31.2861 m/s 350 m We resolve the initial velocity into components. 21 11.2119 m/s 29.2081 m/s Then we find the picklemobiles hang time (which is the same as if it had been shot straight up at about 11.2 m/s), with y = 82 m - 350 m = -268 m. -268 = 11.2119 t - 4.9 t 2 4.9 t 2 - 11.2119 t - 268 = 0 t = -6.3394 s or 8.6276 s (using quadratic formula or computer) continued on next slide 82 m continued on next slide 261 m Slide 16 Picklemobile Example (cont.) 29.2081 m/s continued on next slide We want the positive answer for t. The interpretation of the negative answer is that if the pickle car had been launched from the height of the ledge, it would have taken about 6.3 s to reach the edge of the cliff. Anyway, for 8.62757 s the pickle mobile is in the air and traveling to the right at about 29 m/s. Therefore, its range is (29.2081 m/s) (8.6276 s) 252 m < 261 m. Alas, the poor picklemobile doesnt make it. 82 m Slide 17 Picklemobile Example (cont.) parabolic trajectory 350 m What max height does the pickle mobile attain? 11.2119 m/s It attains the same max height as if it had been shot up at about 11.2 m/s. Since its vertical velocity is zero at its high pt., we have 0 2 - (11.2119) 2 = 2(-9.8) y. So, y = 6.41 m. Add 350 m and the max height is 356.41 m. continued on next slide 82 m Slide 18 83.5805 m/s 29.2081 m/s Picklemobile Example (cont.) What is the impact velocity of the pickle mobile (the velocity upon splash down)? 29.2081 m/s 11.2119 m/s The horiz. component is the same at landing as it was on liftoff. We must find the final vertical velocity: v f 2 - (11.2119) 2 = 2(-9.8) (-350). So, v f = -83.5805 m/s. 350 m The Pythag. theorem gives us the magnitude of the resultant. = tan -1 (83.5805 / 88.5371) = 70.74 . Thus the impact velocity is about 88.5 m/s at 71 below the horizontal. 88.5371 m/s Slide 19 Parabolic Proof A projectile is shot with speed v 0 at an angle . Its vertical position is given by y = (v 0 sin ) t + (-g) t 2. Here y is the dependent quantity, and t is the independent quantity. Everything else is a constant. The projectiles horizontal position is given by x = (v 0 cos ) t. Only x and t are variables, and t = x / (v 0 cos ). Lets substitute this for t in the equation for y : y = (v 0 sin ) t + (-g) t 2 y = (v 0 sin ) [ x / (v 0 cos ) ] - g [ x / (v 0 cos ) ] 2 y = (tan ) x - g 2 v 0 2 cos 2 x 2x 2 The coefficients of x and x 2 are constants. Since the leading coef. is negative, this is the equation of a parabola opening down. Slide 20 Symmetry and Velocity The projectiles speed is the same at points directly across the parabola (at the same vertical position). The angle is the same too, but with opposite orientation. Horizontal speeds are the same throughout the trajectory. Vertical speeds are the same only at points of equal height. The vert. comp. shrinks then grows in opposite direction at a const. rate (- g ). The resultant velocity vectors orientation and magni- tude changes, but is always tangent. The horiz. comp. doesnt change. At the peak, the horiz. comp. equals the resultant velocity vector. Slide 21 Symmetry and Time t = 0 t = 10 t = 20 t = 15 t = 5 t = 3 t = 17 Over level ground, the time at the peak is half the hang time. Notice the symmetry of times at equal heights relative to the 10 unit mark. The projectile has covered half its range when it has peaked, but only over level ground. Note: near the peak the object moves more slowly than when lower to the ground. It rises 3/4 of its max height in only 1/2 of its rising time. (See if you can prove this for an arbitrary launch velocity.) Slide 22 Max height & hang time depend only on initial vertical velocity Each initial velocity vector below has the a different magnitude (speed) but each object will spend the same time in the air and reach the same max height. This is because each vector has the same vertical component. The projectiles will have different ranges, however. The greater the horizo

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