Motion - Physics - Lecture Slides, Slides of Physics

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Motion in Two

Dimensions:

Projectile Motion

Circular Motion

Angular Speed

Simple Harmonic Motion

Torque

Center of Mass

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., they’re falling down at the same rate, and they hit the ground at the same time. Gravity doesn’t care that the black ball is moving sideways; it pulls it downward just the same. Since gravity can’t 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).

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Projectile Motion (cont.)

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t = 0 t = 1

t = 2

t = 3

t = 4

vy = 1

vy = 2

vy = 3

vy = 4

vy = 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 vf = 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 doesn’t.

Since the shot black marble experiences no horiz. forces (ignoring air), it undergoes no horiz. acceleration. Therefore, its horiz. velocity, doesn’t 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 = vx for the shot marble. v = vy for the dropped marble for the whole trip.

Projectile Motion (cont.)

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vx = v vx

vx

vx

vx

v

v

v

vy vy

vy vy

vy

t = 0 t = 1

t = 2

t = 3

t = 4 Docsity.com

x = 1

y = 1^ Projectile Motion^ (cont.)

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 we’ve 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 Docsity.com

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 80 m/s

They hit at same time.

vy vy

Use vf = 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 vy in the pic for each bullet is -2.94 m/s.

Using the Pythagorean theorem for the fired bullet we get 80.

m/s in a direction tangent to its path. continued on next slide

Now let’s 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

v 0

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Step 2: Find hang time. Use y = v 0 t + ½ a t 2 with only vertical data:

v 0 cos

v 0 sin

v 0

y = ( v 0 sin ) t + ½ (- 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. Note: If we had shot the projectile from a 100 m cliff, y would be -100 m.

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Projectiles Fired at an Angle (cont.)

Picklemobile Example

A stuntman drives a picklemobile off a 350 m cliff going 70 mph. The angle of elevation of the cliff is 21. He’s hoping to make it across a 261 m wide river and land on a ledge 82 m high. Does he make it?

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

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Picklemobile Example (cont.)

350 m

We resolve the initial velocity into components.

29.2081 m/s

Then we find the picklemobile’s 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)

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82 m

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261 m

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Picklemobile Example (cont.)

parabolic trajectory

350 m

What max height does the pickle mobile attain?

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

02 - (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

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)?

m/s

The horiz. component is the same at landing as it was on liftoff. We must find the final vertical velocity: vf^2 - (11.2119)^2 = 2(-9.8) (-350).

So, vf = -83.5805 m/s.

350 m

The Pythag. theorem gives us the magnitude of the resultant. = tan-1^ (83.5805 / 88.5371) = 70.. Thus the impact velocity is about 88.5 m/s at 71 below the horizontal.

Symmetry and Velocity

The projectile’s 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 vector’s orientation and magni- tude changes, but is always tangent.

The horiz. comp. doesn’t change. At the peak, the horiz. comp. equals the resultant velocity vector.

Symmetry and Time

t = 0

t = 10

t = 20

t = 5 t = 15

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.)

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