Physics 121, Sections 9, 10, 11, and 12 Lecture 4 Announcements Lectures available on the web (short version) For over-enrollment please go to the Physics office P107 Laboratory sessions start next week Go to my web site www.phys.uconn.edu/~rcote Syllabus + homeworks + lectures, etc. WebAssign: go to www.webassign.net and log in username: first letter of first name plus last name e.g. John Fernando Lachance: jlachance institution: uconn password: PeopleSoft ID # (without the initial “0”) 1

Physics 121, Sections 9, 10, 11, and 12 Lecture 3 Today’s Topics: Homework 1: Due Friday Sept. 6 @ 6:00PM Ch.1: # 4, 10, 14, 19, 31, and 33. Ch.2: # 1, 23, 37, and 63. Chapter 3: Forces and motion along a line Position Velocity: average & instantaneous Acceleration: average & instantaneous Force and acceleration Motion with constant acceleration Falling objects Apparent weight 1

Chapter 3: Position / Displacement Displacement is just change in position. X = xf - xi Displacement is a vector quantity A vector quantity has both magnitude and direction A scalar quantity has only magnitude and no direction 10 meters 15 meters Joe xf xi O x = 5 meters { 2

Average Velocity The average velocity v during the time interval t is defined as the displacement x divided by t.

Instantaneous Velocity The instantaneous velocity v is defined as the limit of the average velocity as the time interval t becomes infinitesimally short

Average Acceleration The average acceleration a during the time interval t, is defined as the change in velocity v divided by the t It is a vector …

Instantaneous Acceleration The instantaneous acceleration a is defined as the limit of the average acceleration as the time interval t goes to zero

Kinematic Variables Measured with respect to a reference frame. (x-y axis) Measured using coordinates (having units). Many kinematic variables are vectors, which means they have a direction as well as a magnitude. Vectors denoted by boldface v or arrow

Newton’s 2nd law and acceleration It relates the net force to the acceleration A force is what changes the velocity of a particle No net force: no change in velocity The constant of proportionality is the mass For a given Fnet a more massive object will undergo a smaller acceleration (and vice versa)

Motion in 1 dimension In 1-D, we usually write position as x . Since it’s in 1-D, all we need to indicate direction is + or . Displacement in a time t = tf - ti is x = xf - xi t x ti tf x t xi xf some particle’s trajectory in 1-D

1-D kinematics Velocity v is the “rate of change of position” Average velocity vav in the time t = tf - ti is: t x t1 t2 x x1 x2 trajectory t Vav = slope of line connecting x1 and x2.

1-D kinematics... Consider limit tf - ti  0 Instantaneous velocity v is defined as: t x t1 t2 x x1 x2 t so v(t2 ) = slope of line tangent to path at t2.

1-D kinematics...  Acceleration a is the “rate of change of velocity” Average acceleration aav in the time t = t2 - t1 is: If a = aav is constant, and set t1=0, v1= v0 , and then v2= v 

Average Velocity Velocity is: v = v0 + at (for constant acceleration) vav v0

1-D kinematics We saw that v = x /  t After a little algebra we have: x = v   t Graphically, we can add up lots of small rectangles: x(t) t + +...+ = displacement

1-D kinematics vav = x /  t For constant acceleration a we find: Recall : Again, set t1=0 and t2=t so that  t = t And so

Recap So for constant acceleration we find: x v t a t

Derivation: Solving for t: Plugging in for t:

Recap: For constant acceleration: From which we know: Active Figure 2

Lecture 4, Act 1 Motion in One Dimension When throwing a ball straight up, which of the following is true about its velocity v and its acceleration a at the highest point in its path? (a) Both v = 0 and a = 0. (b) v  0, but a = 0. (c) v = 0, but a  0. y

Free Fall When any object is let go it falls toward the ground !! The force that causes the objects to fall is called gravity. The acceleration caused by gravity is typically written as g Any object, be it a baseball or an elephant, experiences the same acceleration (g) when it is dropped, thrown, spit, or hurled, i.e. g is a constant.

Gravity facts: g does not depend on the nature of the material! Galileo (1564-1642) figured this out without fancy clocks & rulers! demo - feather & penny in vacuum Nominally, g = 9.81 m/s2 At the equator g = 9.78 m/s2 At the North pole g = 9.83 m/s2 More on gravity in a few lectures!

Lecture 4, Act 2 A car is traveling with an initial velocity vo. At t = 0, the driver puts on the brakes, which slows the car at a rate of ab x = 0, t = 0 ab vo

Lecture 4, Act 2 A car traveling with an initial velocity vo. At t = 0, the driver puts on the brakes, which slows the car at a rate of ab. At what time tf does the car stop, and how much farther xf does it travel ?? v0 ab x = 0, t = 0 v = 0 x = xf , t = tf

Problem: On a bright sunny day you are walking around the campus watching one of the many construction sites. To lift a bunch of bricks from a central area, they have brought in a helicopter. As the pilot is leaving, she accidentally releases the bricks when they are 1000 m above the ground. The worker below is getting ready to walk away in 10 seconds. Does he live?

Problem Solution Method: Five Steps: Focus the Problem - draw a picture – what are we asking for? Describe the physics what physics ideas are applicable what are the relevant variables known and unknown Plan the solution what are the relevant physics equations Execute the plan solve in terms of variables solve in terms of numbers Evaluate the answer are the dimensions and units correct? do the numbers make sense?

We need to find the time it takes for the brick to hit the ground. Problem: 1000 m We need to find the time it takes for the brick to hit the ground.

Problem: Free Fall Constant acceleration at value g Choose coordinate system Variables: h – height of helicopter = 1000 m g – acceleration due to gravity = 9.81 m/s2 t – time to drop – goal m – mass of brick – unknown vo – initial velocity = 0 why? 1000 m y y = 0

Next write position for constant acceleration: Problem: Next write position for constant acceleration: 1000 m Realize that v0y = 0 Solve for when y=0 Why ?? y y = 0

Problem: Substitute y=0 and vo=0 Solve for t in symbols: Solve for t in numbers: y y = 0

Units worked out to be ‘s’ That’s correct for time Problem: Does it make sense? Units worked out to be ‘s’ That’s correct for time It takes 14.3 s. That means that the man escapes !!! y0 = 1000 m y y = 0

Problem #2 As you are driving to school one day, you pass a construction site for a new building and stop to watch for a few minutes. A crane is lifting a batch of bricks on a pallet to an upper floor of the building. Suddenly a brick falls off the rising pallet. You clock the time it takes for the brick to hit the ground at 2.5 seconds. The crane, fortunately, has height markings and you see the brick fell off the pallet at a height of 22 meters above the ground. A falling brick can be dangerous, and you wonder how fast the brick was going when it hit the ground. Since you are taking physics, you quickly calculate the answer. a. Draw a picture illustrating the fall of the brick, the length it falls, and the direction of its acceleration. b. What is the problem asking you to find? c. What kinematics equations will be useful? d. Solve the problem in terms of symbols. e. Does you answer have the correct dimensions? f. Solve the problem with numbers.

Solution / Problem #2 vo g 0= h +v0 t - g t2 / 2 v = v0 - g t h = 22 m brick floor t = 2.5 s g vo a. Draw a diagram b. What do you need to calculate Speed of the brick when it hits the ground (v) c. Which kinematics equations will be useful? acceleration in free fall (a = const = g ~ 10 m/s2) 0= h +v0 t - g t2 / 2 v = v0 - g t

Solution / Problem #2 (cont.) d. Solve the problem in terms of symbols. 1 2 t v0 = ( g t2 - h ) h = v0 t - g t2 / 2 1 2 t v = ( g t2 - h ) - g t v = v0 - g t e. Does you answer have the correct dimensions (what are they)? { } v = 1 2 [ T ] [ L / T 2 ]  [ T ] 2 - [ L ] - [ L / T 2 ]  [ T ] f. Solve the problem with numbers. { } v = 1 2 2.5 s 10 m /s2  (2.5 s )2 - 22 m - 10 m /s2  2.5 s = - 21.3 m /s

Lecture 4, ACT 3 1D Freefall Alice and Bill are standing at the top of a cliff of height H. Both throw a ball with initial speed v0, Alice straight down and Bill straight up. The speed of the balls when they hit the ground are vA and vB respectively. Which of the following is true: vA < vB (b) vA = vB (c) vA > vB v0 Bill Alice H vA vB

Recap of today’s lecture Chapter 3: Forces and motion along a line Position Velocity: average & instantaneous Acceleration: average & instantaneous Force and acceleration Motion with constant acceleration Falling objects Homework 1 on WebAssign 27