Motion in Two or Three Dimensions Chapter 3 Motion in Two or Three Dimensions

Vectors to represent the position of a body Goals for Chapter 3 Vectors to represent the position of a body Velocity vector using the path of a body Acceleration vector of a body Describe the curved path of projectile Investigate circular motion (Describe the velocity of a body as seen from different frames of reference)

Examples of projectile motion. Notice the effects of air resistance

Projectile Motion A projectile is an object moving in two dimensions under the influence of Earth's gravity; its path is a parabola.

Projectile motion Projectile is any body with initial velocity that follows path determined by gravity (& air resistance). Begin by assuming “g” is constant (near ground), & neglecting air resistance, Earth’s curvature, rotation, & motion.

Projectile Motion Understood by analyzing the horizontal and vertical motions separately.

Projectile Motion Photo shows 2 balls starting to fall at the same time. Yellow ball on right has an initial speed in the x-direction. Red ball on left has vx0 = 0 Vertical positions of the balls are identical at identical times Horizontal position of yellow ball increases linearly in time.

Projectile Motion The speed in the x-direction is constant; in the y-direction the object moves with constant acceleration g.

The x and y motion are separable—Figure 3.16 We can analyze projectile motion as horizontal motion with constant velocity and vertical motion with constant acceleration: ax = 0 and ay = g.

Solving Problems Involving Projectile Motion Projectile motion is motion with constant acceleration in two dimensions, Usually where vertical acceleration is g and coordinate system assumes UP = + motion!

Solving Problems Involving Projectile Motion Read the problem carefully, and choose the object(s) you are going to analyze. Draw a diagram. Choose an origin and a coordinate system. Decide on the time interval; this is the same in both directions, and includes only the time the object is moving with constant acceleration g. Examine the x and y motions separately.

Solving Problems Involving Projectile Motion 6. List known and unknown quantities. Remember that vx never changes, and that vy = 0 at the highest point. 7. Plan how you will proceed. Use the appropriate equations; you may have to combine some of them, or find time from one set and use in the other

Solving Problems Involving Projectile Motion Driving off a cliff A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance.

Solving Problems Involving Projectile Motion Example Driving off a cliff. A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance. We WANT: vx (m/s)

Solving Problems Involving Projectile Motion Example Driving off a cliff. A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance. We KNOW: vy0 = 0 m/s!

Solving Problems Involving Projectile Motion Example Driving off a cliff. A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance. We KNOW: Dy = -50 m!

Solving Problems Involving Projectile Motion Example Driving off a cliff. A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance. We KNOW: Dx = +90 m!

Solving Problems Involving Projectile Motion Example Driving off a cliff. A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance. We KNOW: ay = -9.8! ax = 0!

A body projected horizontally A motorcycle leaves a cliff horizontally.

Solving Problems Involving Projectile Motion For Horizontal: Dx = +90 m Dx = vxt + ½ axt2 but ax = 0 so Dx = vxt vx = Dx /t So vx = 90/t How do we get t ??

Solving Problems Involving Projectile Motion Get t from Vertical: Dy = -50 m Dy = vyt + ½ ayt2 but ax = - 9.8 and v0y = 0 so Dy = -50 = - ½ gt2 So t = [-2(-50)/9.8] ½ = 3.2 sec

Solving Problems Involving Projectile Motion Get v0x from time: Dx = +90 m vx = 90/t = 90/3.2 = 28.2 m/sec =~ 62 mile/hour

Projectile Motion If object is launched at initial angle θ0 with the horizontal, analysis is similar except that the initial velocity has a vertical component.

Solving Problems Involving Projectile Motion Example: A kicked football. A football is kicked at an angle θ0 = 37.0° with a velocity of 20.0 m/s, as shown. It travels through the air (assume no resistance) and lands at the ground.

Example: A kicked football. Calculate the maximum height, the time of travel before the football hits the ground, how far away it hits the ground. the velocity vector at the maximum height, the acceleration vector at maximum height.

Solving Problems Involving Projectile Motion STEP 1: Coordinates! (and signs!) +x horizontally to the right +y vertically upwards acceleration of gravity = -9.8 m/s/s

Solving Problems Involving Projectile Motion STEP 2: COMPONENTS! (and signs!) +vx0 horizontally to the right = v0 cos q +vy0 vertically upwards = v0 sinq ay acceleration of gravity = -9.8 m/s/s only in y

Solving Problems Involving Projectile Motion STEP 3: LOOK FOR 3 THINGS! vf = vi + at Dx = ½ (vi + vf)*t Dx = vi*t + ½ at2 Dx = vf*t – ½ at2 Vf2 = Vi2 +2aDx

Solving Problems Involving Projectile Motion In x direction, ax = 0! vfx = vix Dx = vix*t = v cosq * t

Solving Problems Involving Projectile Motion In y direction, ay = -9.8! vfy = viy – 9.8t = v0 sin q - gt Dy = ½ (viy + vfy)*t Dy = viy*t – 4.9t2 = (v0 sin q) t - ½ gt2 Dx = vfy*t + 4.9t2 vfy2 = viy2 – 19.6Dy = (v0 sin q)2 - 2gDy

Solving Problems Involving Projectile Motion

Solving Problems Involving Projectile Motion Solve it! vx = 16 m/s vy = 12 m/s Max height = 7.3 m Max distance in x (“range”) = 39 m Time = 2.5 seconds

Example: Level horizontal range. The horizontal range is defined as the horizontal distance the projectile travels before returning to its original height (which is typically the ground); that is, y(final) = y0. (a) Derive a formula for the horizontal range R of a projectile in terms of its initial speed v0 and angle θ0.

Solving Problems Involving Projectile Motion Example: Level horizontal range (b) Suppose one of Napoleon’s cannons had a muzzle speed, v0, of 60.0 m/s. At what angle should it have been aimed (ignore air resistance) to strike a target 320 m away?

Solving Problems Involving Projectile Motion Example: A kicked football. A football is kicked at an angle θ0 = 37.0° with a velocity of 20.0 m/s, as shown. It travels through the air (assume no resistance) and lands at the ground.

Solving Problems Involving Projectile Motion A football is kicked at an angle θ0 = 37.0° with a velocity of 20.0 m/s, as shown. Suppose the football was punted and left the punter’s foot at a height of 1.00 m above the ground. How far did the football travel before hitting the ground?

Solving Problems Involving Projectile Motion Suppose the football was punted and left the punter’s foot at a height of 1.00 m above the ground. How far did the football travel before hitting the ground? Dy = - 1 meter = vyt + ½ ayt2

A good problem to try algebraically… 3.61 - A projectile is being launched from ground level with no air resistance. You want to avoid having it enter a temperature inversion layer in the atmosphere a height h above the ground. What is the maximum launch speed you could give this projectile if you shot it straight up? Express your answer in terms of h and g . Suppose the launcher available shoots projectiles at twice the maximum launch speed you found in part A. At what maximum angle above the horizontal should you launch the projectile?

Solving Problems Involving Projectile Motion The wrong strategy! A boy on a small hill aims his water-balloon slingshot horizontally, straight at a second boy hanging from a tree branch a distance d away. At the instant the water balloon is released, the second boy lets go and falls from the tree, hoping to avoid being hit. Show that he made the wrong move. (He hadn’t studied physics yet.) Ignore air resistance.

Tranquilizing a falling monkey Where should the zookeeper aim?

Solving Problems Involving Projectile Motion Projectile motion is parabolic: Taking the equations for x and y as a function of time, and combining them to eliminate t, we find y as a function of x: This is the equation for a parabola.

The equations for projectile motion If we set x0 = y0 = 0, the equations describing projectile motion are shown at the right. The trajectory is a parabola.

Height and range of a projectile A baseball is batted at an angle. Vo = 37.0 m/s at 53.1 degrees; @ 2 seconds where is it?

Maximum height and range of a projectile What initial angle will give the maximum height and the maximum range of a projectile?

The effects of air resistance Calculations become more complicated. Acceleration is not constant. Effects can be very large. Maximum height and range decrease. Trajectory is no longer a parabola.

Position vector The position vector from the origin to point P has components x, y, and z.

Average velocity Average velocity between two points = displacement divided by time Same direction as the displacement.

Instantaneous velocity Instantaneous velocity is instantaneous rate of change of position vector with respect to time. Components of instantaneous velocity are vx = dx/dt, vy = dy/dt, and vz = dz/dt. Instantaneous velocity of a particle is always tangent to its path.

Average acceleration The average acceleration during a time interval t is defined as the velocity change during t divided by t.

Average acceleration The average acceleration during a time interval t is defined as the velocity change during t divided by t.

Instantaneous acceleration Instantaneous acceleration is the instantaneous rate of change of the velocity with respect to time. Any particle following a curved path is accelerating, even if it has constant speed. Components are ax = dvx/dt, ay = dvy/dt, and az = dvz/dt.

Direction of the acceleration vector Direction of acceleration vector depends on whether speed is constant, increasing, or decreasing,

Direction of the acceleration vector Direction of acceleration vector depends on whether speed is constant, increasing, or decreasing,

Uniform circular motion For uniform circular motion, the speed is constant and the acceleration is perpendicular to the velocity.

Acceleration for uniform circular motion For uniform circular motion, the instantaneous acceleration always points toward the center of the circle and is called the centripetal acceleration. The magnitude of the acceleration is arad = v2/R. The period T is the time for one revolution, and arad = 4π2R/T2.

Nonuniform circular motion If the speed varies, the motion is nonuniform circular motion. The radial acceleration component is still arad = v2/R, but there is also a tangential acceleration component atan that is parallel to the instantaneous velocity.

Acceleration of a skier Skier moving on a ski-jump ramp. Figure shows the direction of the skier’s acceleration at various points.

Relative velocity The velocity of a moving body seen by a particular observer is called the velocity relative to that observer, or simply the relative velocity. A frame of reference is a coordinate system plus a time scale.

Relative velocity in one dimension If point P is moving relative to reference frame A, we denote the velocity of P relative to frame A as vP/A. If P is moving relative to frame B and frame B is moving relative to frame A, then the x-velocity of P relative to frame A is vP/A-x = vP/B-x + vB/A-x.

Relative velocity on a straight road Motion along a straight road is a case of one-dimensional motion.

Relative velocity in two or three dimensions We extend relative velocity to two or three dimensions by using vector addition to combine velocities. A passenger’s motion is viewed in the frame of the train and the cyclist.

Flying in a crosswind A crosswind affects the motion of an airplane.