Motion in Two and Three Dimensions Chapter 4

Position and Displacement A position vector locates a particle in space o Extends from a reference point (origin) to the particle Example o Position vector (-3m, 2m, 5m) Eq. (4-1) Figure 4-1

Position and Displacement Change in position vector is a displacement We can rewrite this as: Or express it in terms of changes in each coordinate: Eq. (4-2) Eq. (4-3) Eq. (4-4)

Average Velocity and Instantaneous Velocity Average velocity is o A displacement divided by its time interval We can write this in component form: Example o A particle moves through displacement (12 m)i + (3.0 m)k in 2.0 s: Eq. (4-8) Eq. (4-9)

Average Velocity and Instantaneous Velocity Instantaneous velocity is o The velocity of a particle at a single point in time o The limit of avg. velocity as the time interval shrinks to 0 Visualize displacement and instantaneous velocity: Eq. (4-10) Figure 4-4 Figure 4-3

Average Velocity and Instantaneous Velocity In unit-vector form, we write: Which can also be written: Note: a velocity vector does not extend from one point to another, only shows direction and magnitude Eq. (4-11) Eq. (4-12)

Average Acceleration and Instantaneous Acceleration Average acceleration is o A change in velocity divided by its time interval Instantaneous acceleration is again the limit t → 0: We can write Eq in unit-vector form: Eq. (4-15) Eq. (4-16)

Average Acceleration and Instantaneous Acceleration We can rewrite as: To get the components of acceleration, we differentiate the components of velocity with respect to time Eq. (4-17) Eq. (4-18) Figure 4-6

Average Acceleration and Instantaneous Acceleration Note: as with velocity, an acceleration vector does not extend from one point to another, only shows direction and magnitude Answer: (1) x:yes, y:yes, a:yes (3) x:yes, y:yes, a:yes (2) x:no, y:yes, a:no (4) x:no, y:yes, a:no

Two-dimensional kinematics is similar to one-dimensional kinematics, but we must now use full vector notation rather than positive and negative signs to indicate the direction of motion.

Characteristics of Projectile Motion: Horizontal Motion of a ball rolling freely along a level surface. Horizontal velocity is ALWAYS constant. Vertical Motion of a freely falling object. Force due to gravity. Vertical component of velocity changes with time. Parabolic Path traced by an object accelerating only in the vertical direction while moving at constant horizontal velocity.  The most important experimental fact about projectile motion in two dimensions is that the horizontal and vertical motions are completely independent of each other. This means that motion in one direction has no effect on motion in the other direction.  In general, the equations of constant acceleration developed in previous lectures follow separately for both the x-direction and the y-direction. An important difference is that the initial velocity now has two components. Projectile Motion

A projectile is o A particle moving in the vertical plane o With some initial velocity o Whose acceleration is always free-fall acceleration (g) The motion of a projectile is projectile motion Launched with an initial velocity v 0 Eq. (4-19) Eq. (4-20)

Projectile Motion Therefore we can decompose two-dimensional motion into 2 one-dimensional problems Figure 4-10 Answer: Yes. The y-velocity is negative, so the ball is now falling.

Projectile Motion Horizontal motion: o No acceleration, so velocity is constant (recall Eq. 2-15): V Vertical motion: o Acceleration is always -g (recall Eqs. 2-15, 2-11, 2-16): Eq. (4-21) Eq. (4-22) Eq. (4-23) Eq. (4-24)

Projectile Motion

Projectile motion EXAMPLE 2 : A ball is thrown horizontally with an initial velocity of 25 m/s from the top of a 60 m high building. a) how much time will the ball take to hit the ground ? b) how far does the ball travel horizontally ? c) what is the velocity of the ball as it hits the ground ? © 2014 John Wiley & Sons, Inc. All rights reserved.

a) Y = Y 0 +(V 0 sin ѳ )t -1/2gt 2 Ѳ = 0 Y= -1/2gt 2 t= 3.5s b) X=(v 0 cos ѳ )t X=87.5 m c) v x =v 0 cos Ѳ =25 m/s v y =v 0 sin Ѳ – gt = m/s ן v ן= m/s © 2014 John Wiley & Sons, Inc. All rights reserved.

Projectile motion Example 3 : a projectile is fired to achieve a maximum range of 120 m.what should its speed be ? R MAX =V 2 0 /g V 0 =37 m/s Example 4 : a projectile is fired with an angle Θ above the horizontal. take 15s to reach its range of 110 m. What is its speed at the highest point ? x=(v 0 cos Ѳ )t v 0 cos Ѳ = v 0x = x/t = 9.34 m/s © 2014 John Wiley & Sons, Inc. All rights reserved.

Projectile Motion HW # 4 A ball is thrown up with initial velocity of 15 m/s at an angle of (30) from the top of a 50 m high building. It lands on the ground at a point P. 1- Calculate the maximum height the ball can reach ? 2- how long does it take the ball to hit the ground at p ? 3- how far from the building edge does the ball hit the ground ? 4- what is the speed of the ball just before hitting the ground ? © 2014 John Wiley & Sons, Inc. All rights reserved.

Uniform Circular Motion A particle is in uniform circular motion if o It travels around a circle or circular arc o At a constant speed Since the velocity changes, the particle is accelerating! Velocity and acceleration have: o Constant magnitude o Changing direction Figure 4-16

Uniform Constant speed, or, constant magnitude of velocity. Circular motion along a circle: Changing direction of velocity. Centripetal Acceleration: centripetal means center-seeking. تسارع الجاذبية : يعني مركز التماس. Uniform Circular Motion

Acceleration is called centripetal acceleration o Means “center seeking” o Directed radially inward The period of revolution is: o The time it takes for the particle go around the closed path exactly once Eq. (4-34) Eq. (4-35)

Uniform Circular Motion

Position Vector Locates a particle in 3-space Displacement Change in position vector Average and Instantaneous Accel. Average and Instantaneous Velocity Eq. (4-2) Eq. (4-8) Eq. (4-15) Summary Eq. (4-1) Eq. (4-3) Eq. (4-4) Eq. (4-10) Eq. (4-16)

Projectile Motion Flight of particle subject only to free-fall acceleration (g) Uniform Circular Motion Magnitude of acceleration: Time to complete a circle: Eq. (4-34) Summary Eq. (4-22) Eq. (4-23) Eq. (4-35)

Exercise 1, 3, 6, 13, 56, 62 © 2014 John Wiley & Sons, Inc. All rights reserved.