Physics 2514Lecture 22

P. Gutierrez

Department of Physics & AstronomyUniversity of Oklahoma

Physics 2514 – p. 1/15

Information

Information needed for the examExam will be in the same format as the practice with thesame number of questionsBring a # 2 pencil & eraserCalculators will be allowedNo cell phones, no laptops, . . .Only exam, pencil, eraser, calculator allowed on desk.Bring student id with you

You will need to knowStudent id numberDiscussion section #Your name

Physics 2514 – p. 2/15

Exam 2

Exam 2 will cover material in lectures 11 through 21. Thisincludes material in Chapters 5 through 8

KinematicsProjectile motion & general 2-d motion with constantacceleration;Circular motion;Relative motion.

DynamicsNewton’s 3 laws of motion.

Physics 2514 – p. 3/15

Formulas to be given

The following formulas will be given

s(t) =1

2at2 + v0t + s0 v(t) = at + v0

v2 − v20 = 2a(s − s0) s = rθ

~v = ~v′ + ~V ~Fnet = m~a

fk = µn fs ≤ µn

Physics 2514 – p. 4/15

Review Kinematics

Started with discussion of motionDisplacement final minus initial position, with directionpointing from initial to final position

~r = ∆~r = ~rf −~ri

Velocity is the rate of change in position

~vavg =∆~r

∆t⇒ ~v(t) =

d~r

dt

Acceleration is rate at which the velocity changes

~aavg =∆~v

∆t⇒ ~a(t) =

d~v

dt

Physics 2514 – p. 5/15

Review 2-d Kinematics

Kinematic equations for constant acceleration (Applyindependently in each dimension)

Positions(t) =

1

2at2 + v0t + s0

Velocityv(t) = at + v0

Combining equations

v2 − v2

0= 2a(s − s0)

Find constraint that ties equations togetherHow long does it take to fall, how far does it travel horizontally

Physics 2514 – p. 6/15

Steps in Problem Solving

Steps in problem solvingA) Rewrite the problem eliminating all extraneous

information. (What are you given, what are you looking,what are the constraints);

B) Draw a diagram along with a coordinate system, labeleach object with the variables associated with it;

C) What are the known and unknown quantities, whichunknowns are you solving for;

D) Write down the equations associated with the problem,and solve the problem algebraically (SIMPLIFY!!!)

E) Finally, substitute numbers into the equation, andcalculate the numerical solution

Physics 2514 – p. 7/15

Example

A catapult is tested by Roman legionnaires. They tabulate theresults in a papyrus scroll and 2000 years later anarchaeological team reads (distance translated into modernunits): Range = 0.20 km; angle of launch = π/3. What is theinitial velocity of launch of the boulder?

Object launched at angle π/3 radians, travels 0.2 km. Determineinitial velocity using the constraint, how long does it take to hitthe ground?PSfrag replacements

v0v0

vx

vyθθ

Physics 2514 – p. 8/15

Example

Object launched at angle π/3 radians, travels 0.2 km. Determine initialvelocity using the constraint, how long it take to hit the ground?

PSfrag replacements

v0v0

vx

vyθθ

yf = −1

2gt2f + (v0 sin θ)tf

xf = (v0 cos θ)tf

⇒

tf =

√

2xf tan θi − 2yf

g= 8.4 s

v0 =xf

√

gxf tan θi − gyf√2(xf sin θ − yf cos θ)

= 47.6 m/s

xf ≡ x(tf ) = 200 m, yf ≡ y(tf ) = 0 m, θi ≡ θ(ti) = π/3 radsPhysics 2514 – p. 9/15

Circular Motion

Arc-Length & Angle

s = rθ

Tangential & Angular Velocity

v =ds

dt= r

dθ

dt= rω

Uniform Motion |~v| = constant

s(t) = v0t + s0 ⇒ θ = ω0t + θ0

v(t) = v0 ⇒ ω(t) = ω0

Physics 2514 – p. 10/15

Nonuniform Circular Motion

Consider that case when the speed around a circle changes.

Physics 2514 – p. 11/15

Kinematic Equations

The kinematic equations for uniform circular motion werederived earlier, here we consider nonuniform motionMotion along arc is 1-D with tangentialacceleration and velocity determining mo-tion

s = s0 + vott +1

2att

2

vt = v0t + att

Now divide by r

1

r(s = s0 + v0tt +

1

2att

2)

1

r(vt = v0t + att)

θ = θ0 + ω0t +1

2αt2

ω = ω0 + αt α = at/r

Physics 2514 – p. 12/15

Centripetal Acceleration

Calculate average acceleration

CB = ∆~r2−∆~r1 = ~v2∆t−~v1∆t = ∆~v∆t

Angles

ABO: θ + α + α = 180

DAC: φ + α + α = 180

)

⇒ θ = φ

Similar triangles

CBAB

=ABAO

⇒|∆~v|∆t

v∆t=

v∆t

r

Average radial acceleration

aaverager =

|∆~v|

∆t=

v2

r

ar = lim∆t→0

|∆~v|

∆t=

v2

r

Physics 2514 – p. 13/15

Example

A car starts from rest on a curve with a radius of 120 m andaccelerates at 1.0 m/s2. Through what angle will the car have traveledwhen the magnitude of its total acceleration is 2.0 m/s2.

PSfrag replacements

vt

at

ar

Knowns

at = 1.0 m/s2

af =√

a2t + a2

r = 2.0 m/s2

vt0 = 0 m/sr = 120 mθ0 = 0 rads

Unknowns

θf = ? rads

Physics 2514 – p. 14/15

Example

A car starts from rest on a curve with a radius of 120 m andaccelerates at 1.0 m/s2. Through what angle will the car have traveledwhen the magnitude of its total acceleration is 2.0 m/s2.Knowns

at = 1.0 m/s2

af =√

a2t + a2

r = 2.0 m/s2

vt0 = 0 m/sr = 120 mθ0 = 0 rads

Unknowns

θf = ? rads

How long to reach af

af =

√

a2t +

(

v2

tf

r

)2

= 2.0 m/s2

vtf = 14.4 m/svtf = attf ⇒ tf = 14.4 s

How farθf = 1

2αt2f = 1

2

at

rt2f ⇒ θf = 0.864 rads

Physics 2514 – p. 15/15