Physics 1200 Mechanics, Kinematics, Fluids, Waves • Lecturer: Tom Humanic
• Contact info: Office: Physics Research Building, Rm. 2144 Email: [email protected] Phone: 614 247 8950
• Office hours: Tuesday 3:00 pm, Wednesday 11:00 am
My lecture slides may be found on my website at http://www.physics.ohio-state.edu/~humanic/ Chapter 1
Measurement, units, … Units SI units
meter (m): unit of length
second (s): unit of time Units The Role of Units in Problem Solving
1 ft = 0.3048 m
1 mi = 1.609 km
1 hp = 746 W
1 liter = 10-3 m3 The Role of Units in Problem Solving
Example: Interstate Speed Limit
Express the speed limit of 65 miles/hour in terms of meters/second.
Use 5280 feet = 1 mile and 3600 seconds = 1 hour and 3.281 feet = 1 meter.
& miles # & miles #& 5280 feet #& 1 hour # feet Speed = $65 !(1)(1)= $65 !$ !$ != 95 % hour " % hour "% mile "% 3600 s " second
& feet # & feet #& 1 meter # meters Speed = $95 !(1)= $95 !$ ! = 29 % second " % second "% 3.281 feet " second Scalars and Vectors
A scalar quantity is one that can be described by a single number:
temperature, speed, mass
A vector quantity deals inherently with both magnitude and direction:
Chapter 2
Kinematics in One Dimension Displacement 1 dimensional motion: the car can only travel either to the left or right
-x +x
xo = initial position vector x = final position vector Δx = x − xo = displacement vector Displacement
xo = 2.0 m Δx = 5.0 m
x = 7.0 m
Δx = x − xo = 7.0 m − 2.0 m = 5.0 m Speed and Velocity
Average speed is the distance traveled divided by the time required to cover the distance.
Distance Average speed = Elapsed time
SI units for speed: meters per second (m/s) Speed and Velocity
Example: Distance Run by a Jogger
How far does a jogger run in 1.5 hours (5400 s) if his average speed is 2.22 m/s?
Distance Average speed = Elapsed time
Distance = (Average speed)(Elapsed time) = (2.22 m s)(5400 s)=12000 m Speed and Velocity
Average velocity is the displacement divided by the elapsed time. Displacement Average velocity = Elapsed time
x − x Δx v = o = t − to Δt Speed and Velocity
Example: The World’s Fastest Jet-Engine Car
Andy Green in the car ThrustSSC set a world record of 341.1 m/s in 1997. To establish such a record, the driver makes two runs through the course, one in each direction, to nullify wind effects. From the data, determine the average velocity for each run.
Δx +1609 m v = = = +339.5m s Δt 4.740 s
Δx −1609 m v = = = −342.7 m s Δt 4.695 s Speed and Velocity
The instantaneous velocity indicates how fast the car moves and the direction of motion at each instant of time.
Δx v = lim Δt→0 Δt Acceleration
DEFINITION OF AVERAGE ACCELERATION v − vo Δv a = = SI unit: m/s2 t − to Δt
DEFINITION OF INSTANTANEOUS ACCELERATION Δv a = lim Δt→0 Δt Acceleration
Example: Acceleration and Increasing Velocity
Determine the average acceleration of the plane.
v 0 km h o = v = 260km h to = 0 s t = 29 s
v − v 260km h − 0km h km h a = o = = +9.0 t − to 29 s − 0 s s Acceleration Equations of Kinematics for Constant Acceleration x − xo v − v v = a = o t − t o t − to
For 1-dimensional motion it is customary to dispense with the use of boldface symbols overdrawn with arrows for the displacement, velocity, and acceleration vectors. We will, however, continue to convey the directions with a plus or minus sign.
_ x − x _ v − v v = o a = o t − to t − to Equations of Kinematics for Constant Acceleration
Let the object be at the origin when the clock starts.
xo = 0 to = 0
x − x x v = o v = t t t − o True for constant acceleration
1 x = vt = 2 (vo + v)t Equations of Kinematics for Constant Acceleration True for constant acceleration v − v v − v a = a = o a = o t − to t
at = v − vo
v = vo + at Equations of Kinematics for Constant Acceleration
Five kinematic variables:
1. displacement, x
2. acceleration (constant), a
3. final velocity (at time t), v
4. initial velocity, vo
5. elapsed time, t Equations of Kinematics for Constant Acceleration
v = vo + at
1 1 x = 2 (vo + v)t = 2 (vo + vo + at)t
1 2 x = vot + 2 at Equations of Kinematics for Constant Acceleration
Boat moving with constant acceleration --> find x
1 2 x = vot + 2 at 1 2 2 = (6.0m s)(8.0 s)+ 2 (2.0m s )(8.0 s) = +110 m Equations of Kinematics for Constant Acceleration
Example: Catapulting a Jet
Find its displacement. 2 vo = 0m s a = +31m s x = ?? v = +62m s Equations of Kinematics for Constant Acceleration
v − vo v − v a = t = o t a
1 1 (v − vo ) x = (vo + v)t = (vo + v) 2 2 a
v2 − v2 x = o 2a Equations of Kinematics for Constant Acceleration
2 2 2 2 v − vo (62m s) − (0m s) x = = = +62 m 2a 2(31m s2 ) Equations of Kinematics for Constant Acceleration
Equations of Kinematics for Constant Acceleration
v = vo + at
1 x = 2 (vo + v)t
2 2 v = vo + 2ax
1 2 x = vot + 2 at Applications of the Equations of Kinematics Reasoning Strategy 1. Make a drawing.
2. Decide which directions are to be called positive (+) and negative (-).
3. Write down the values that are given for any of the five kinematic variables.
4. Verify that the information contains values for at least three of the five kinematic variables. Select the appropriate equation.
5. When the motion is divided into segments, remember that the final velocity of one segment is the initial velocity for the next.
6. Keep in mind that there may be two possible answers to a kinematics problem. Applications of the Equations of Kinematics
Example: An Accelerating Spacecraft A spacecraft is traveling with a velocity of +3250 m/s. Suddenly the retrorockets are fired, and the spacecraft begins to slow down with an acceleration whose magnitude is 10.0 m/s2. What is the velocity of the spacecraft when the displacement of the craft is +215 km, relative to the point where the retrorockets began firing?
x a v vo t
+215000 m -10.0 m/s2 ? +3250 m/s Applications of the Equations of Kinematics
x a v vo t
+215000 m -10.0 m/s2 ? +3250 m/s
2 2 2 v = vo + 2ax v = ± vo + 2ax
v = ± (3250m s)2 _+ 2(10.0m s2 )(215000 m) = ±2500m s Applications of the Equations of Kinematics Freely Falling Bodies
In the absence of air resistance, it is found that all bodies at the same location above the Earth fall vertically with the same acceleration. If the distance of the fall is small compared to the radius of the Earth, then the acceleration remains essentially constant throughout the descent.
This idealized motion is called free-fall and the acceleration of a freely falling body is called the acceleration due to gravity.
g = 9.80m s2 or 32.2ft s2 Freely Falling Bodies
g = 9.80m s2 Freely Falling Bodies
Example: A Falling Stone
A stone is dropped from the top of a tall building. After 3.00s of free fall, what is the displacement y of the stone? Freely Falling Bodies
y a v vo t ? -9.80 m/s2 0 m/s 3.00 s Freely Falling Bodies
y a v vo t ? -9.80 m/s2 0 m/s 3.00 s
1 2 y = vot + 2 at 1 2 2 = (0m s)(3.00 s)+ 2 (− 9.80m s )(3.00 s) = −44.1 m Freely Falling Bodies
Example: How High Does it Go?
The referee tosses the coin up with an initial speed of 5.00m/s. In the absence if air resistance, how high does the coin go above its point of release? Freely Falling Bodies
y a v vo t ? -9.80 m/s2 0 m/s +5.00 m/s Freely Falling Bodies
y a v vo t ? -9.80 m/s2 0 m/s +5.00 m/s
2 2 2 2 v − vo v = vo + 2ay y = 2a
2 2 2 2 v − vo (0m s) − (5.00m s) y = = =1.28 m 2a 2(− 9.80m s2 ) Freely Falling Bodies
Conceptual Example: Acceleration Versus Velocity
There are three parts to the motion of the coin. On the way up, the coin has a vector velocity that is directed upward and has decreasing magnitude. At the top of its path, the coin momentarily has zero velocity. On the way down, the coin has downward-pointing velocity with an increasing magnitude.
In the absence of air resistance, does the acceleration of the coin, like the velocity, change from one part to another? Graphical Analysis of Velocity and Acceleration Object moving with constant velocity (zero acceleration)
Δx + 8 m Slope = = = +4m s = v Δt 2 s Graphical Analysis of Velocity and Acceleration Changing velocity during a bike trip
v = 0 m/s
v = -1 m/s v = 2 m/s Graphical Analysis of Velocity and Acceleration Object moving with changing velocity
Slope of the tangent line is the instantaneous velocity at the t = 20 s point: Slope = Δx/Δt = (26 m)/(5 s) = 5.2 m/s = v Graphical Analysis of Velocity and Acceleration Object moving with constant acceleration
Δv +12 m s 2 Slope = = = +6m s = a Δt 2 s