Wednesday, September 24, 2014 Vectors Review

Announcements Lab fee? Matt, Emily Spangler, Morgan, Brittany Woods, Giovanni? Lab Safety Contract Wyatt HW:

Scalars vs Vectors Scalars have magnitude only Distance, speed, time, mass Vectors have both magnitude and direction displacement, velocity, acceleration R  head tail

Direction of Vectors The direction of a vector is represented by the direction in which the ray points. This is typically given by an angle. A x 

Magnitude of Vectors The magnitude of a vector is the size of whatever the vector represents. The magnitude is represented by the length of the vector. Symbolically, the magnitude is often represented as │ A │ A If vector A represents a displacement of three miles to the north… B Then vector B, which is twice as long, would represent a displacement of six miles to the north!

Equal Vectors Equal vectors have the same length and direction, and represent the same quantity (such as force or velocity).

Inverse Vectors Inverse vectors have the same length, but opposite direction. A -A

A B R A + B = R Graphical Addition of Vectors Vectors are added graphically together head-to- tail. The sum is called the resultant. The inverse of the sum is called the equilibrant

Component Addition of Vectors 1) Resolve each vector into its x- and y- components. A x = Acos  A y = Asin  B x = Bcos  B y = Bsin  etc. 2) Add the x-components together to get R x and the y-components to get R y. 3) Use the Pythagorean Theorem to get the magnitude of the resultant. 4) Use the inverse tangent function to get the angle.

Sample problem: Add together the following graphically and by component, giving the magnitude and direction of the resultant and the equilibrant. –Vector A: o –Vector B: o –Vector C: o

Thursday, August 28, 2008 Unit Vectors

Friday, August 29, 2008 Position, Velocity, and Acceleration Vectors in Multiple Dimensions

1 Dimension2 or 3 Dimensions x: position  x: displacement v: velocity a: acceleration r: position  r: displacement v: velocity a: acceleration r = x i + y j + z k  r =  x i +  y j +  z k v = v x i + v y j + v z k a = a x i + a y j + a z k In Unit Vector Notation

Sample problem: The position of a particle is given by r = (80 + 2t)i – 40j - 5t 2 k. Derive the velocity and acceleration vectors for this particle. What does motion “look like”?

Sample problem: A position function has the form r = x i + y j with x = t 3 – 6 and y = 5t - 3. a) Determine the velocity and acceleration functions. b) Determine the velocity and speed at 2 seconds.

Miscellaneous Let ’ s look at some video analysis. Let ’ s look at a documentary. Homework questions?

Tuesday, September 2, 2008 Multi-Dimensional Motion with Constant (or Uniform) Acceleration

Sample Problem: A baseball outfielder throws a long ball. The components of the position are x = (30 t) m and y = (10 t – 4.9t 2 ) m a) Write vector expressions for the ball’s position, velocity, and acceleration as functions of time. Use unit vector notation! b) Write vector expressions for the ball’s position, velocity, and acceleration at 2.0 seconds.

Sample problem: A particle undergoing constant acceleration changes from a velocity of 4i – 3j to a velocity of 5i + j in 4.0 seconds. What is the acceleration of the particle during this time period? What is its displacement during this time period?

Trajectory of Projectile g g g g g This shows the parabolic trajectory of a projectile fired over level ground. Acceleration points down at 9.8 m/s 2 for the entire trajectory.

Trajectory of Projectile vxvx vyvy vyvy vxvx vxvx vyvy vxvx vyvy vxvx The velocity can be resolved into components all along its path. Horizontal velocity remains constant; vertical velocity is accelerated.

Position graphs for 2-D projectiles. Assume projectile fired over level ground. x y t y t x

t VyVy t VxVx Velocity graphs for 2-D projectiles. Assume projectile fired over level ground.

Acceleration graphs for 2-D projectiles. Assume projectile fired over level ground. t ayay t axax

Remember… To work projectile problems… …resolve the initial velocity into components.  VoVo V o,y = V o sin  V o,x = V o cos 

Sample problem: A soccer player kicks a ball at 15 m/s at an angle of 35 o above the horizontal over level ground. How far horizontally will the ball travel until it strikes the ground?

Sample problem: A cannon is fired at a 15 o angle above the horizontal from the top of a 120 m high cliff. How long will it take the cannonball to strike the plane below the cliff? How far from the base of the cliff will it strike?

Sample problem: derive the trajectory equation.

Sample problem: Derive the range equation for a projectile fired over level ground.

Sample problem: Show that maximum range is obtained for a firing angle of 45 o.

Wednesday, September 3, 2008 Monkey Gun

Announcements Homework policy change.

Will the projectile always hit the target presuming it has enough range? The target will begin to fall as soon as the projectile leaves the gun.

Friday, September 5, 2008 Review of Uniform Circular Motion

Uniform Circular Motion Occurs when an object moves in a circle without changing speed. Despite the constant speed, the object ’ s velocity vector is continually changing; therefore, the object must be accelerating. The acceleration vector is pointed toward the center of the circle in which the object is moving, and is referred to as centripetal acceleration.

Vectors in Uniform Circular Motion a v a = v2 v2 / r v a v a v a

Sample Problem The Moon revolves around the Earth every 27.3 days. The radius of the orbit is 382,000,000 m. What is the magnitude and direction of the acceleration of the Moon relative to Earth?

Sample problem: Space Shuttle astronauts typically experience accelerations of 1.4 g during takeoff. What is the rotation rate, in rps, required to give an astronaut a centripetal acceleration equal to this in a simulator moving in a 10.0 m circle.

Wednesday, September 10, 2008 Radial and Tangential Acceleration

Tangential acceleration Sometimes the speed of an object in circular motion is not constant (in other words, it’s not uniform circular motion). An acceleration component is tangent to the path, aligned with the velocity. This is called tangential acceleration. The centripetal acceleration component causes the object to continue to turn as the tangential component causes the radius or speed to change.

v Tangential Acceleration radial or centripetal component (a r or a c ) tangential component (a T ) a If tangential acceleration exists, the orbit is not stable.

Sample Problem: Given the figure at right rotating at constant radius, find the radial and tangential acceleration components if  = 30 o and a has a magnitude of 15.0 m/s 2. What is the speed of the particle? How is it behaving? 5.00 m a 

Sample problem: Suppose you attach a ball to a 60 cm long string and swing it in a vertical circle. The speed of the ball is 4.30 m/s at the highest point and 6.50 m/s at the lowest point. Find the acceleration of the ball at the highest and lowest points.

Sample problem: A car is rounding a curve on the interstate, slowing from 30 m/s to 22 m/s in 7.0 seconds. The radius of the curve is 30 meters. What is the acceleration of the car?

Thursday, September 11, 2008 Relative Motion

When observers are moving at constant velocity relative to each other, we have a case of relative motion. The moving observers can agree about some things, but not about everything, regarding an object they are both observing.

Consider two observers and a particle. Suppose observer B is moving relative to observer A. P particle A observer B v rel

Also suppose particle P is also moving relative to observer A. P particle A observer B v rel vAvA In this case, it looks to A like P is moving to the right at twice the speed that B is moving in the same direction.

However, from the perspective of observer B… P particle A observer B -v rel vBvB it looks like P is moving to the right at the same speed that A is moving in the opposite direction, and this speed is half of what A reports for P. vAvA v rel

The velocity measured by two observers depends upon the observers’ velocity relative to each other. P particle A observer B -v rel vBvB v B = v A – v rel v A = v B + v rel vAvA v rel

Sample problem: Now show that although velocity of the observers is different, the acceleration they measure for a third particle is the same provided v rel is constant. Begin with v B = v A - v rel

Galileo’s Law of Transformation of Velocities If observers are moving but not accelerating relative to each other, they agree on a third object ’ s acceleration, but not its velocity!

Inertial Reference Frames Frames of reference which may move relative to each other but in which observers find the same value for the acceleration of a third moving particle. Inertial reference frames are moving at constant velocity relative to each other. It is impossible to identify which one may be at rest. Newton ’ s Laws hold only in inertial reference frames, and do not hold in reference frames which are accelerating.

Sample problem: How long does it take an automobile traveling in the left lane at 60.0km/h to pull alongside a car traveling in the right lane at 40.0 km/h if the cars’ front bumpers are initially 100 m apart?

Sample problem: A pilot of an airplane notes that the compass indicates a heading due west. The airplane’s speed relative to the air is 150 km/h. If there is a wind of 30.0 km/h toward the north, find the velocity of the airplane relative to the ground.