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Welcome back to Physics 215. Today’s agenda: Velocity and acceleration in two-dimensional motion Motion under gravity -- projectile motion Acceleration on curved path. Current homework assignment. HW2: Ch.2 (Knight textbook): 48, 78, 82 Ch.3 (Knight textbook): 28, 32, 42

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Welcome back to physics 215

Welcome back to Physics 215

Today’s agenda:

Velocity and acceleration in two-dimensional motion

Motion under gravity -- projectile motion

Acceleration on curved path


Current homework assignment
Current homework assignment

HW2:

  • Ch.2 (Knight textbook): 48, 78, 82

  • Ch.3 (Knight textbook): 28, 32, 42

  • Ch.4 (Knight textbook): 40

  • due Wednesday, Sept 15th in recitation

Reminder about course website:

http://www.phy.syr.edu/courses/PHY215.10Fall/index.html


Exam 1 next thursday 9 23 10
Exam 1: next Thursday (9/23/10)

  • In room 208 (here!) at the usual lecture time

  • Material covered:

    • Textbook chapters 1 - 4

    • Lectures up through 9/21 (slides online)

    • Wed/Fri Workshop activities

    • Homework assignments

  • Work through practice exam problems (posted on website)

  • Work on more practice exam problems next Wednesday in recitation workshop


Displacement in 2d motion
Displacement in 2D Motion

y

Ds

sI

sF

s – vector position

Displacement s = sF - sI, also a vector!

O

x


2d motion in components
2D Motion in components

  • x and y motions decouple

  • vx = dx/dt vy = dy/dt

  • ax = dvx/dt ay = dvy/dt

  • If acceleration is only non-zero in 1 direction, can choose coordinates so that 1 component of acceleration is zero

    • e.g., motion under gravity


Motion under gravity
Motion under gravity

ax = 0

vx = v0x

x= x0 + v0xt

ay = -g

vy = v0y - gt

y= y0 + v0yt - (1/2)gt2

y

v0y = v0sin(q)

v0x = v0cos(q)

v0

q

x

Projectile motion...


Welcome back to physics 215

A ball is ejected vertically upward from a cart at rest. The ball goes up, reaches its highest point and returns to the cart.

In a second experiment, the cart is moving at constant velocity and the ball is ejected in the same way, where will the ball land?

  • In front of the cart.

  • B. Behind the cart.

  • C. Inside the cart.

  • D. The outcome depends on the speed of the cart.


Projectile question
Projectile question ball goes up, reaches its highest point and returns to the cart.

  • A ball is thrown at 45o to vertical with a speed of 7 m/s. Assuming g=10 m/s2, how far away does the ball land?


Welcome back to physics 215

A battleship simultaneously fires two shells at enemy ships. If the shells follow the parabolic trajectories shown, which ship will be hit first?

A. A

B. Both at the same time

C. B

D. need more information


Projectile motion
Projectile motion If the shells follow the parabolic trajectories shown, which ship will be hit first?

y

x

R : when is y=0 ?

t[vy1-(1/2)gt] = 0

i.e., T = (2v)sinq/g

 R (x-eqn.)

 hmax (y-eqn.)


Maximum height and range
Maximum height and range If the shells follow the parabolic trajectories shown, which ship will be hit first?


Motion on a curved path at constant speed
Motion on a curved path If the shells follow the parabolic trajectories shown, which ship will be hit first?at constant speed

Is the acceleration of the object equal to zero?


Velocity is tangent to path
Velocity is tangent to path If the shells follow the parabolic trajectories shown, which ship will be hit first?

Ds

sI

sF

O

v = Ds/Dt lies along

dotted line. As Dt  0

direction of v is tangent

to path


Motion on a curved path at constant speed1
Motion on a curved path If the shells follow the parabolic trajectories shown, which ship will be hit first?at constant speed


Subtracting vectors
Subtracting vectors If the shells follow the parabolic trajectories shown, which ship will be hit first?

Recall that vF+ (-vI) = Dv

Dv

vF

vI

same as

-vI

vF

Dv


For an object moving at constant speed along a curved path the acceleration is not zero

For an object moving at constant If the shells follow the parabolic trajectories shown, which ship will be hit first?speed along a curved path, the acceleration is not zero.


Welcome back to physics 215

For which of the following motions of a car does the change in velocity vector have the greatest magnitude? (All motions occur at the same constant speed.)

A. A 90° right turn at constant speed

B. A U-turn at constant speed

C. A 270° turn on a highway on-ramp

D. The change in velocity is zero for all three motions.


Welcome back to physics 215

A car moves along the path shown. Velocity vectors at two different points are sketched. Which of the arrows below most closely represents the direction of the average acceleration?

A. B.

C. D.


Welcome back to physics 215

A child is riding a bicycle on a level street. The velocity and acceleration vectors of the child at a given time are shown.

Which of the following velocity vectors may represent the velocity at a later time?

a

A. B.

C. D.


Welcome back to physics 215

A biker is riding at constant speed clockwise on the oval track shown below.

Which vector correctly describes the acceleration at the point indicated?


Welcome back to physics 215

Biker moving around oval at track shown below.constant speed

As point D is moved closer to C, angle approaches 90°.


Summary
Summary track shown below.

  • For motion at constant speed, instantaneous acceleration vector is perpendicular to velocity vector

  • Points ``inward’’

  • What is the magnitude of the acceleration vector?


Welcome back to physics 215

Acceleration vectors for ball swung in a horizontal circle at constant speed v

v1

v2

R

q

v1

q

v2

What is the magnitude of the acceleration?a = v2/R


Demo shooting the bear
DEMO: shooting the bear … at

  • Bear released at same time ball projected from end of tube

  • What happens to the bear and the ball?

  • Does outcome depend on angle relative to floor?


Reading assignment
Reading assignment at

  • Circular motion

  • 4.5 – 4.7 in textbook