Physics 231 Topic 3: Vectors and two dimensional motion Alex Brown September 14-18 2015

What’s up? (Monday Sept 14) 1) Homework set 01 due Tuesday Sept 15th 10 pm 2) Learning Resource Center (LRC) in 1248 BPS Monday 9 am to 1 pm and 3 pm to 5 pm Tuesday 9 am to 9 pm Friday 9am to 1 pm

Key Concepts: 2D Motion Vectors and Scalars Two Dimensional Motion Velocity in 2D Acceleration in 2D Projectile motion Throwing a ball or cannon fire Uniform Circular Motion Centripetal acceleration Covers chapter 3 in Rex & Wolfson

Extra Credit Quiz Impossible to tell -9.8 m/s2 9.8 m/s2 You throw a ball straight up (positive direction) and catch it again at the same location where you released it. At the highest point the acceleration is: Impossible to tell -9.8 m/s2 9.8 m/s2

Trigonometry  SOH – CAH - TOA: sin =opposite/hypotenuse =a/c cos =adjacent/hypotenuse =b/c tan =opposite/adjacent =a/b c Pythagorean theorem: a Note that sin,cos,tan are dimensionless. 2 radians corresponds to 360o  b

Be Careful…. h y y   y = h sin() = h cos() Always check carefully which angle is given

Vectors and Scalars Scalar: A Vector: A or A (bold face) Scalar: A quantity specified by its magnitude only Vector: A quantity specified both by its magnitude and direction. To distinguish a vector from a scalar quantity, it is usually written with an arrow above it, or in bold to distinguish it from a scalar. Scalar: A Vector: A or A (bold face)

Vectors and Scalars

Question No Are these two vectors the same? Yes Are the lengths of these two vectors the same? No Yes Two vectors are equal if both their lengths and directions are the same!

Vector Addition A+B B A B+A A B A+B=B+A

Vector operations in equations (xa+b,ya+b) y (xb,yb) A+B B (xa,ya) A x Example: A+B

Vector Subtraction B A A-B = A + (-B) -B

Vector Subtraction -B A-B=A+(-B) A B

More on Coordinate Systems Cartesian Coordinates: (x,y)=(a,b) Also, r = a î + b ĵ î unit vector in x direction ĵ unit vector in y direction a b r  x Plane Polar Coordinates (r,)

Vector length and its components Y (xa,ya)  x Length of vector (use pythagorean theorem):

Vector operations in equations

Vector operations in equations ff

Question A man walks 5 km/h. He travels 12 minutes to the east, 30 minutes to the south-east and 36 minutes to the north. A) What is the displacement of the man when he’s done? B) What is the total distance he walked? A) 1 km 2.5 km 3 km x2=2.5 cos()=1.77 =-45o Y2=2.5 sin()=-1.77 B) 1 + 2.5 + 3 = 6.5 km

Boat crossing the river Flow

Question A boat is trying to cross a 1-km wide river in the shortest way (straight across). Its maximum speed (in still water) is 10 km/h. The river is flowing (south) with 5 km/h. 1) At what angle  does the captain have to steer the boat the go straight across? 2) How long does it take for the boat to cross the river? 

Answer  sin = opposite/hypotenuse = 5/10 = 0.5  = sin-1(0.5) = 30o A boat is trying to cross a 1-km wide river in the shortest way (straight across). Its maximum speed (in still water) is 10 km/h. The river is flowing with 5 km/h. At what angle  does the captain have to steer the boat the go straight across? 2) How long does it take the boat to cross the river? v=10 km/h  sin = opposite/hypotenuse = 5/10 = 0.5  = sin-1(0.5) = 30o Flow=5km/h velocityhor=sqrt(100-25) = 8.66 km/h Time = (1 km)/(8.66 km/h) = 0.115 h = 6.9 min

3) If it doesn’t matter at what point the boat reaches the other side, at what angle should the captain steer to cross in the fastest way? 0o :the horizontal component of the velocity is then maximum.

plane in the wind

Displacement in 2D ff

Velocity in 2D

1d motion 2d motion; decompose into horizontal and vertical components

So for motion in x-direction Near the surface of the earth with the y-axis pointing up ax = 0 ay = -g So for motion in x-direction and for motion in y direction the same equations (A-E) as before but using y (rather than x)

(A) (B) (C) (D) (E) positive y points up so a = -g

Pop and Drop

Pop and Drop B A For A: vy = - ½gt2 vx = 0 y = y0- ½gt2 x = 0 For B: vy = - ½gt2 vx = v0 y = y0 – ½gt2 x =x0t

Clicker Quiz Galileo and Newton stand on top of the tower of Pisa. Galileo drops a stone of mass 2 kg straight down (no initial velocity). Newton throws a 2 kg stone with an initial horizontal velocity of 3 m/s. Which stone will hit the ground first? (ignore effects of friction) a) The stone thrown by Galileo b) The stone thrown by Newton c) Both stones arrive at the same time d) Not enough information to say.

Clicker Quiz Galileo and Newton stand on top of the tower of Pisa. Galileo drops a stone of mass 2 kg straight down (no initial velocity). Newton throws a 2 kg stone with an initial horizontal velocity of 3 m/s. Which stone will hit the ground first? (ignore effects of friction) a) The stone thrown by Galileo b) The stone thrown by Newton c) Both stones arrive at the same time d) Not enough information to say.

Clicker Quiz Firing Balls I A cart is rolling at constant velocity on a flat track. It fires a ball straight up into the air as it moves. After it is fired, what happens to the ball? a) it depends on how fast the cart is moving b) it falls behind the cart c) it falls in front of the cart d) it falls right back into the cart

Clicker Quiz Firing Balls I A small cart is rolling at constant velocity on a flat track. It fires a ball straight up into the air as it moves. After it is fired, what happens to the ball? a) it depends on how fast the cart is moving b) it falls behind the cart c) it falls in front of the cart d) it falls right back into the cart In the frame of reference of the cart, the ball only has a vertical component of velocity. So it goes up and comes back down. To a ground observer, both the cart and the ball have the same horizontal velocity, so the ball still returns into the cart. when viewed from cart when viewed from ground

Relative motion Motion is relative to a reference frame! Motion of the ball in rest-frame of cart Resulting motion Motion of the cart

When studying motion in 2D it is often convenient to decomposition the motion into horizontal and vertical components, both of which are both described in 1D Remember that the object can accelerate in one direction, but remain at the same speed in the other direction. Remember that after decomposition of 2D motion into horizontal and vertical components, you should investigate both components to understand the complete motion of a particle. After decomposition into horizontal and vertical directions, treat the two directions independently – but the time is common to both.

Parabolic motion

Vx remains constant throughout the flight! Parabolic motion vx(t) = v0x vy (t) = v0y – g t vx=v0 cos vy=v0 sin-2g=0 vx=v0 cos vy=v0 sin-1g vx=v0 cos vy=v0 sin-3g v = (vx, vy) vx=v0 cos vy=v0 sin Vx remains constant throughout the flight! vx=v0 cos vy=v0 sin-4g  t=0 t=1 t=2 t=3 t=4

Extra Credit Quiz Impossible to tell -9.8 m/s2 9.8 m/s2 You throw a ball in an arch (“up” is the positive y direction), at the highest point in the arc the y component of the acceleration is: Impossible to tell -9.8 m/s2 9.8 m/s2

Parabolic motion Where is the speed… …highest?  t=0 t=1 t=2 t=3 E C D

Parabolic motion Where is the speed… …lowest ?  t=0 t=1 t=2 t=3 E C D

Question A hunter aims at a bird that is some distance away and flying very high (i.e. consider the vertical position of the hunter to be 0), but he misses. If the bullet leaves the gun with a speed of v0 and friction by air is negligible, with what speed vf does the bullet hit the ground after completing its parabolic path? v0 vf

Answer First consider the horizontal direction: v0x = v0cos() Since there is no friction, there is no change in the horizontal component: vx (tf) = v0x Next the vertical direction with equations A and B: vy(t) = v0y - gt y(t) = voy t - ½ gt2 (g = 9.8 m/s2) When the bullet hits the ground y(tf)=0: 0 = voytf - ½gtf2 tf = 0 or tf = 2v0y/g So, vy(tf) = v0y - g(2v0y/g) = - v0y Total speed = v0x2 + (-v0y)2 = v0!!!! The speed of the bullet has not changed, but the vertical component of the velocity has changed sign.

Clicker Quiz Dropping a Package You drop a package from a plane flying at constant speed in a straight line. Without air resistance, the package will: a) quickly lag behind the plane while falling b) remain vertically under the plane while falling c) move ahead of the plane while falling d) not fall at all

What does the motion of the object look like according to the pilot?

Clicker Quiz Dropping a Package You drop a package from a plane flying at constant speed in a straight line. Without air resistance, the package will: a) quickly lag behind the plane while falling b) remain vertically under the plane while falling c) move ahead of the plane while falling d) not fall at all Both the plane and the package have the same horizontal velocity at the moment of release. They will maintain this velocity in the x-direction, so they stay aligned.

Another example (A) (B) (C) (D) (E) x(t) = v0 cos()t y(t) = v0 sin()t – ½gt2 (A) A football player throws a ball with initial velocity of 30 m/s at an angle of 30o degrees w.r.t. the ground. 1) How far will the ball fly before hitting the ground? 2) What about an angle of 60o? 3) At which angle is the distance thrown maximum? y(t) = 0 if… t( v0 sin() - ½gt )=0 tf=0 or tf = 2 v0 sin()/g (B) (C) x(tf) = v0 cos() tf = 2 (v0 )2 cos() sin() / g if  = 30o x = 79.5 m if  = 60o x = 79.5 m !! (D) Maximum if cos() sin() is maximum, so =45o x(=45o) = 91.7 m (E)

And another example…

And another example… Vertical direction y(t) = y0 + v0 sin()t - ½gt2 11 = 37 + 10.1 sin(59)t - ½9.8t2 4.9t2 - 8.66t - 26=0 t = 3.35 (solve quadratic equation) This time correspond to how long it takes for the ball to land on the second building.

And another example… Vertical direction y(t) = y0 + v0 sin()t - ½gt2 Calculate how far the ball goes in the horizontal direction. Vertical direction y(t) = y0 + v0 sin()t - ½gt2 11 = 37 + 10.1 sin(59)t - ½9.8t2 4.9t2 - 8.66t - 26=0 t = 3.35 (solve quadratic equation) This time correspond to how long it takes for the ball to land on the second building. Horizontal direction x(t) = x0 + v0 cos() t use time derived from vertical direction x(3.35)=0 + 10.1 cos(59) 3.35 x(3.35)=17.4 m

Shoot the monkey The hunter aims his gun exactly at the monkey At the moment the hunter fires, the monkey drops off the branch. What happens? The hunter aims his gun exactly at the monkey At the moment he fires, the monkey drops off the branch. What happens? monkey gets hit bullet goes over the monkey bullet goes under the monley no idea PHY 231

The horizontal position of the bullet is: x(t) = v0 cos() t when x(t) = d then td = d/[v0 cos()] and its vertical position is: y(t) = v0 sin() t – ½gt2 y(td) = d tan() – ½gt2 = h - ½gt2 y=h v0  y=0 X=0 X=d The vertical position of the monkey is: y(t) = h – ½gt2 The horizontal position is d PHY 231

Uniform circular motion with constant speed Consider a car at constant speed, constrained to a circular track Velocity is always determined by the direction the car is facing at a given time Velocity is constantly changing, so this implies an acceleration But it’s just the direction of the velocity that’s changing! r

Uniform circular motion with constant speed v

Angular Frequency f = 1/T  = 2 π f Earth orbits the sun with a period of 365 days. T=365 days(24 h/day)(3600 s/h) = 3.15107 sec Frequency: f = 1/T = 1/(3.15107 sec) = 3.1710-8 Hz R  Earth travels at constant speed throughout its orbit: v = x/t It must traverse the circumference of the orbit: c = 2 π R Thus, the speed v = c/T = 2 π R / T We can also express this in terms of an angular frequency: The angular frequency  =  /  t = 2 π / T  = the speed at which the angle is changing Units = radians/s f = 1/T  = 2 π f

Clicker Question A hummingbird with mass 4g flies northwest at a speed of 10 m/s and at an angle of 30o relative to the horizontal. It beats its wings once every 0.014 sec. Assuming it flies at constant velocity, what is the beat frequency of its wings? 0.14 Hz 7.1 Hz 14 Hz 71 Hz None of the above f = 1/T T = 0.014 sec f = 1/0.014 sec Frequency: f = 71 Hz

Uniform circular motion with constant speed v

 

 

𝑎 = 𝑣 𝑓 − 𝑣 𝑖 𝑡 𝑓 − 𝑡 𝑖 = Δ𝑣 Δ𝑡 Sin( /2 ) = ( s/2 ) / r Sin( /2 ) = ( v/2 ) / v

𝑎 𝑐 =𝑣 𝑣 𝑟 = 𝑣 2 𝑟 𝑎 = 𝑣 𝑓 − 𝑣 𝑖 𝑡 𝑓 − 𝑡 𝑖 = Δ𝑣 Δ𝑡 𝑎 = 𝑣 𝑓 − 𝑣 𝑖 𝑡 𝑓 − 𝑡 𝑖 = Δ𝑣 Δ𝑡 ( s/2 ) / r = sin( /2 ) sin( /2 ) = ( v/2 ) / v ( s/2 ) / r = ( v/2 ) / v ( s/t ) v/r = ( v/t ) ( s/t ) = v 𝑎 𝑐 =𝑣 𝑣 𝑟 = 𝑣 2 𝑟

Uniform circular motion with constant speed Which way does the centripetal acceleration point ?

Uniform circular motion with constant speed

𝑎 = 𝑣 𝑓 − 𝑣 𝑖 𝑡 𝑓 − 𝑡 𝑖 = Δ𝑣 Δ𝑡 sin( /2 ) = ( s/2 ) / r sin( /2 ) = ( v/2 ) / v

𝑎 𝑐 =𝑣 𝑣 𝑟 = 𝑣 2 𝑟 𝑎 = 𝑣 𝑓 − 𝑣 𝑖 𝑡 𝑓 − 𝑡 𝑖 = Δ𝑣 Δ𝑡 𝑎 = 𝑣 𝑓 − 𝑣 𝑖 𝑡 𝑓 − 𝑡 𝑖 = Δ𝑣 Δ𝑡 ( s/2 ) / r = sin( /2 ) sin( /2 ) = ( v/2 ) / v ( s/2 ) / r = ( v/2 ) / v ( s/t ) v/r = ( v/t ) ( s/t )  v 𝑎 𝑐 =𝑣 𝑣 𝑟 = 𝑣 2 𝑟

Clicker Question begin Which route is shorter? Red Black The same Don’t know end