Physics 111: Mechanics Lectures 4 & 5 PDF

Physics 111: Mechanics Lectures 4 and 5 Sharafadeen Adeniji NUN Physics Department Motion along a straight line  Motion  Position and displacement  Average velocity and average speed  Instantaneous velocity and speed  Acceleration  Constant acceleration: A spec...

Physics 111: Mechanics Lectures 4 and 5 Sharafadeen Adeniji NUN Physics Department Motion along a straight line  Motion  Position and displacement  Average velocity and average speed  Instantaneous velocity and speed  Acceleration  Constant acceleration: A special case  Free fall acceleration Jan. 28-Feb. 1, 2013  Motion  Everything moves! Motion is one of the main topics in Physics  In the spirit of taking things apart for study, then putting them back together, we will LAX first consider only motion along a straight line. Newark  Simplification: Consider a moving object as a particle, i.e. it moves like a particle —a “point object” Jan. 28-Feb. 1, 2013 4 Basic Quantities in Kinematics Jan. 28-Feb. 1, 2013 One Dimensional Position x  Motion can be defined as the change of position over time.  How can we represent position along a straight line?  Position definition: Defines a starting point: origin (x = 0), x relative to origin Direction: positive (right or up), negative (left or down) It depends on time: t = 0 (start clock), x(t=0) does not have to be zero.  x = + 2.5 m Position has units of [Length]: meters. x=-3m Jan. 28-Feb. 1, 2013  Vector and Scalar  A vector quantity is characterized by having both a magnitude and a direction. Displacement, Velocity, Acceleration, Force … Denoted in  boldface type v , a , F... or with an arrow over  the vtop. , a , F...  A scalar quantity has magnitude, but no direction. Distance, Mass, Temperature, Time …  For motion along a straight line, the direction is represented simply by + and – signs.  sign: Right or Up.  sign: Left or Down.  1-D motion can be thought of as a component of 2-D and 3-D motions. Jan. 28-Feb. 1, 2013  Quantities in Motion  Any motion involves three concepts Displacement Velocity Acceleration  These concepts can be used to study objects in motion. Jan. 28-Feb. 1, 2013  Displacement  Displacement is a change of position in time.  Displacement: x x f (t f )  xi (ti ) f stands for final and i stands for initial.  It is a vector quantity.  It has both magnitude and direction:  or  sign  It has units of [length]: meters.x (t ) = + 2.5 m 1 1 x2 (t2) = - 2.0 m Δx = -2.0 m - 2.5 m = -4.5 m x1 (t1) = - 3.0 m x2 (t2) = + 1.0 m Δx = +1.0 m + 3.0 m = +4.0 m Jan. 28-Feb. 1, 2013  Distance and Position-time graph  Displacement in space From A to B: Δx = xB – xA = 52 m – 30 m = 22 m From A to C: Δx = xc – xA = 38 m – 30 m = 8 m  Distance is the length of a path followed by a particle from A to B: d = |xB – xA| = |52 m – 30 m| = 22 m from A to C: d = |xB – xA|+ |xC – xB| = 22 m + |38 m – 52 m| = 36 m  Displacement is not Distance. Jan. 28-Feb. 1, 2013  Velocity  Velocity is the rate of change of position.  Velocity is a vector quantity. displacement  Velocity has both magnitude and direction.  Velocity has a unit of [length/time]: meter/second. distance  We will be concerned with three quantities, defined as: Average velocity x x f  xi vavg   t t Average speed total distance savg  t Instantaneous  x dx v  lim  velocity t 0  t dt displacement Jan. 28-Feb. 1, 2013 Average Velocity  Average velocity x x f  xi vavg   t t is the slope of the line segment between end points on a graph.  Dimensions: length/time (L/T) [m/s].  SI unit: m/s.  It is a vector (i.e. is signed), and displacement direction sets its sign. Jan. 28-Feb. 1, 2013 Average Speed  Average speed total distance savg  t  Dimension: length/time, [m/s].  Scalar: No direction involved.  Not necessarily close to Vavg: Savg = (6m + 6m)/(3s+3s) = 2 m/s Vavg = (0 m)/(3s+3s) = 0 m/s Jan. 28-Feb. 1, 2013 Jan. 28-Feb. 1, 2013 Jan. 28-Feb. 1, 2013 Jan. 28-Feb. 1, 2013 Jan. 28-Feb. 1, 2013  Graphical Interpretation of Velocity  Velocity can be determined from a position-time graph  Average velocity equals the slope of the line joining the initial and final positions. It is a vector quantity.  An object moving with a constant velocity will have a graph that is a straight line. Jan. 28-Feb. 1, 2013  Instantaneous Velocity  Instantaneous means “at some given instant”. The instantaneous velocity indicates what is happening at every point of time.  Limiting process: Chords approach the tangent as Δt => 0 Slope measure rate of change of position  x dx  Instantaneous velocity:v  lim  t 0  t dt  It is a vector quantity.  Dimension: length/time (L/T), [m/s].  It is the slope of the tangent line to x(t).  Instantaneous velocity v(t) is a function of time. Jan. 28-Feb. 1, 2013  Uniform Velocity  Uniform velocity is the special case of constant velocity  In this case, instantaneous velocities are always the same, all the instantaneous velocities will also equal the x x f  xi velocity average x f  xi  v x t vx    Begin with t t then Note: we are x v plotting velocity vs. x(t) time v(t) xf vx xi 0 t 0 t ti tf Jan. 28-Feb. 1, 2013  Average Acceleration  Changing velocity (non-uniform) means an acceleration is present.  Acceleration is the rate of change of velocity.  Acceleration is a vector quantity.  Acceleration has both magnitude and direction.  Acceleration has a dimensions of length/time 2: [m/s2]. v v f  vi aavg    Definition: t t f  ti Average acceleration v dv d dx d 2 v a lim    2 Instantaneous acceleration t 0 t dt dt dt dt Jan. 28-Feb. 1, 2013  Average Acceleration Note: we are plotting velocity vs.  Average acceleration time v v f  vi aavg   t t f  ti  Velocity as a function of time v f (t ) vi  aavg t  It is tempting to call a negative acceleration a “deceleration,” but note: When the sign of the velocity and the acceleration are the same (either positive or negative), then the speed is increasing When the sign of the velocity and the acceleration are in the opposite directions, the speed is decreasing  Average acceleration is the slope of the line connecting the initial and final velocities on a velocity-time graph Jan. 28-Feb. 1, 2013  Instantaneous and Uniform Acceleration  The limit of the average acceleration as the time interval goes to zero v dv d dx d 2 v a lim    t 0 t dt dt dt dt 2  When the instantaneous accelerations are always the same, the acceleration will be uniform. The instantaneous acceleration will be equal to the average acceleration  Instantaneous acceleration is the slope of the tangent to the curve of the velocity-time graph Jan. 28-Feb. 1, 2013 Relationship between Acceleration and Velocity (First Stage)  Velocity and acceleration are in the same direction  Acceleration is uniform (blue arrows maintain the same length)  Velocity is increasing (red arrows are getting longer) v f (t ) vi  at  Positive velocity and positive acceleration Jan. 28-Feb. 1, 2013 Relationship between Acceleration and Velocity (Second Stage)  Uniform velocity (shown by red arrows maintaining the same size)  Acceleration equals v f (t ) vi  at zero Jan. 28-Feb. 1, 2013 Relationship between Acceleration and Velocity (Third Stage)  Acceleration and velocity are in opposite directions  Acceleration is uniform (blue arrows maintain the same length)  Velocity is decreasing (red arrows are getting shorter) v f (t ) vi  at  Velocity is positive and acceleration is negative Jan. 28-Feb. 1, 2013  Kinematic Variables: x, v, a  Position is a function of time: x  x(t )  Velocity is the rate of change of position.  Acceleration is the rate of change of velocity.  x dx v dv v  lim  a  lim  t 0  t dt t 0 t dt d d dt dt  Position Velocity Acceleration  Graphical relationship between x, v, and a This same plot can apply to an elevator that is initially stationary, then moves upward, and then stops. Plot v and a as a function of time. Jan. 28-Feb. 1, 2013 Special Case: Motion with Uniform Acceleration (our typical case)  Acceleration is a constant  Kinematic Equations (which we will derive in a moment) v v0  at 1 x v t  (v0  v)t 2 x v0t  12 at 2 2 2 v v0  2ax Jan. 28-Feb. 1, 2013 Derivation of the Equation (1)  Given initial conditions: a(t) = constant = a, v(t = 0) = v , x(t = 0) = x 0 0  Start with definition of average acceleration: v v  v0 v  v0 v  v0 aavg     a t t  t0 t 0 t  We immediately get the first equation  v v0  at Shows velocity as a function of acceleration and time  Use when you don’t know and aren’t asked to find the displacement Jan. 28-Feb. 1, 2013 Derivation of the Equation (2)  Given initial conditions: a(t) = constant = a, v(t = 0) = v0, x(t = 0) = x0  Start with definition of average velocity: x  x0 x vavg    t t Since velocity changes at a constant rate, we have 1 x vavg t  (v0  v)t  Gives displacement as a2function of velocity and time  Use when you don’t know and aren’t asked for the acceleration Jan. 28-Feb. 1, 2013 Derivation of the Equation (3)  Given initial conditions: a(t) = constant = a, v(t = 0) = v , x(t = 0) = x 0 0  Start with the two just-derived equations: v v0  at x vavg t  1 (v0  v)t 2  We have x  1 (v0  v)t  1 (v0  v0  at )t 1 2 x  x  x0 v0t  at 2 2 2  Gives displacement as a function of all three quantities: time, initial velocity and acceleration  Use when you don’t know and aren’t asked to find the final velocity Jan. 28-Feb. 1, 2013 Derivation of the Equation (4)  Given initial conditions: a(t) = constant = a, v(t = 0) = v , x(t = 0) = x 0 0  Rearrange the definition of average acceleration v v  v , to find the time 0 v  v0 aavg   a t  t t Use it to eliminate t in the second equation:a , rearrange to get 2 2 1 1 v  v0 x  (v0  v)t  (v  v0 )(v  v0 )  2 2a 2a 2 2  v 2 vof Gives velocity as a function 0 2 a x v0  2a ( x  x0 ) acceleration    and displacement  Use when you don’t know and aren’t asked for the time Jan. 28-Feb. 1, 2013  Problem-Solving Hints  Read the problem  Draw a diagram Choose a coordinate system, label initial and final points, indicate a positive direction for velocities and accelerations  Label all quantities, be sure all the units are consistent Convert if necessary v v0  at  Choose the appropriate kinematic equation  Solve for the unknowns x v0t  12 at 2 You may have to solve two equations for two unknowns 2 v2 v0  2ax  Check your results Jan. 28-Feb. 1, 2013  Example  An airplane has a lift-off speed of 30 m/s after a take-off run of 300 m, what minimum constant acceleration? 2 v v0  at x v0t  12 at 2 v 2 v0  2ax  What is the corresponding take-off time? v v0  at x v0t  12 at 2 2 v 2 v0  2ax Jan. 28-Feb. 1, 2013 Jan. 28-Feb. 1, 2013 Free Fall Acceleration Jan. 28-Feb. 1, 2013 Free Fall Acceleration Jan. 28-Feb. 1, 2013 Free Fall Acceleration y  Earth gravity provides a constant acceleration. Most important case of constant acceleration.  Free-fall acceleration is independent of mass.  Magnitude: |a| = g = 9.8 m/s2  Direction: always downward, so ag is negative if we define “up” as positive, a = g = 9.8 m/s2  Try to pick origin so that xi = 0 Jan. 28-Feb. 1, 2013  Free Fall for  Rookie A stone is thrown from the top of a building with an initial velocity of 20.0 m/s straight upward, at an initial height of 50.0 m above the ground. The stone just misses the edge of the roof on the its way down. Determine  (a) the time needed for the stone to reach its maximum height.  (b) the maximum height.  (c) the time needed for the stone to return to the height from which it was thrown and the velocity of the stone at that instant.  (d) the time needed for the stone to reach the ground  (e) the velocity and position of the stone at t = 5.00s Jan. 28-Feb. 1, 2013 Jan. 28-Feb. 1, 2013 Example 2 Jan. 28-Feb. 1, 2013  Velocity and Position by Integration  How can we find the position and velocity in straight-line motion from the acceleration function ax Jan. 28-Feb. 1, 2013 Velocity and Position by Integration Jan. 28-Feb. 1, 2013 Velocity and Position by Integration Jan. 28-Feb. 1, 2013 Example Jan. 28-Feb. 1, 2013  Summary  This is the simplest type of motion  It lays the groundwork for more complex motion  Kinematic variables in one dimension Position x(t) m L Velocity v(t) m/s L/T Acceleration a(t) m/s2 L/T2 All depend on time All are vectors: magnitude and direction vector:  Equations for motion with constant acceleration: missing quantities x – x0 v v0  at x  x0 v0t  12 at 2 v 2 2 vt v0  2a ( x  x0 ) x  x0  12 (v  v0 )t a vx0  x0 vt  1 2 at 2 Jan. 28-Feb. 1, 2013