Physics 111: Lecture 4 Today’s Agenda

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Physics 111: Lecture 4 Today’s Agenda. Recap of centripetal acceleration Newton’s 3 laws How and why do objects move? Dynamics. Review: Centripetal Acceleration. UCM results in acceleration: Magnitude : a = v 2 / R =   R Direction : - r (toward center of circle). ^. v =  R. - PowerPoint PPT Presentation

Transcript of Physics 111: Lecture 4 Today’s Agenda

Physics 106P: Lecture 5 Notes

Physics 111: Lecture 4

Todays AgendaRecap of centripetal accelerationNewtons 3 laws

How and why do objects move? DynamicsPhysics 111: Lecture 4, Pg #Review: Centripetal AccelerationUCM results in acceleration:Magnitude:a = v2 / R = R Direction:- r (toward center of circle)Ra^Useful stuff:

f = rotations / secT = 1 / f = 2 / T = 2 f = rad/secv = RPhysics 111: Lecture 4, Pg #

Physics 111: Lecture 4, Pg #DynamicsIsaac Newton (1643 - 1727) published Principia Mathematica in 1687. In this work, he proposed three laws of motion:

Law 1: An object subject to no external forces is at rest or moves with a constant velocity if viewed from an inertial reference frame.Law 2: For any object, FNET = F = maLaw 3: Forces occur in pairs: FA ,B = - FB ,A (For every action there is an equal and opposite reaction.)Physics 111: Lecture 4, Pg #Newtons First LawAn object subject to no external forces is at rest or moves with a constant velocity if viewed from an inertial reference frame.If no forces act, there is no acceleration.

The following statements can be thought of as the definition of inertial reference frames.An IRF is a reference frame that is not accelerating (or rotating) with respect to the fixed stars.If one IRF exists, infinitely many exist since they are related by any arbitrary constant velocity vector!

1. Dishes2. MonkeyPhysics 111: Lecture 4, Pg #Is Urbana a good IRF?Is Urbana accelerating?YES!Urbana is on the Earth.The Earth is rotating.

What is the centripetal acceleration of Urbana?T = 1 day = 8.64 x 104 sec, R ~ RE = 6.4 x 106 meters .

Plug this in: aU = .034 m/s2 ( ~ 1/300 g)Close enough to 0 that we will ignore it.Urbana is a pretty good IRF.

IcepuckPhysics 111: Lecture 4, Pg #Newtons Second LawFor any object, FNET = F = ma.The acceleration a of an object is proportional to the net force FNET acting on it.The constant of proportionality is called mass, denoted m.This is the definition of mass.The mass of an object is a constant property of thatobject, and is independent of external influences.

Force has units of [M]x[L / T2] = kg m/s2 = N (Newton)

Physics 111: Lecture 4, Pg #Newtons Second Law...What is a force?A Force is a push or a pull.A Force has magnitude & direction (vector).Adding forces is like adding vectors.F1 F2 aF1 F2 aFNET FNET = maPhysics 111: Lecture 4, Pg #Newtons Second Law...Components of F = ma : FX = maX FY = maY FZ = maZ Suppose we know m and FX , we can solve for aX and apply the things we learned about kinematics over the last few weeks:

Physics 111: Lecture 4, Pg #Example: Pushing a Box on Ice.A skater is pushing a heavy box (mass m = 100 kg) across a sheet of ice (horizontal & frictionless). He applies a force of 50 N in the i direction. If the box starts at rest, what is its speed v after being pushed a distance d = 10 m?Fv = 0maiPhysics 111: Lecture 4, Pg #Example: Pushing a Box on Ice.A skater is pushing a heavy box (mass m = 100 kg) across a sheet of ice (horizontal & frictionless). He applies a force of 50 N in the i direction. If the box starts at rest, what is its speed v after being pushed a distance d = 10m ?dFvmaiPhysics 111: Lecture 4, Pg #Example: Pushing a Box on Ice...Start with F = ma.a = F / m.Recall that v2 - v02 = 2a(x - x0 )(Lecture 1)

So v2 = 2Fd / m

dFvmaiPhysics 111: Lecture 4, Pg #Example: Pushing a Box on Ice...Plug in F = 50 N, d = 10 m, m = 100 kg:Find v = 3.2 m/s.dFvmai

Physics 111: Lecture 4, Pg #Lecture 4, Act 1Force and accelerationA force F acting on a mass m1 results in an acceleration a1.The same force acting on a different mass m2 results in an acceleration a2 = 2a1.If m1 and m2 are glued together and the same force F acts on this combination, what is the resulting acceleration?(a) 2/3 a1 (b) 3/2 a1 (c) 3/4 a1F a1m1F a2 = 2a1m2F a = ? m1m2Physics 111: Lecture 4, Pg #Lecture 4, Act 1Force and accelerationSince a2 = (1/2) a1 for the same applied force, m2 = (1/2)m1 !m1+ m2 = 3m2 /2(a) 2/3 a1 (b) 3/2 a1 (c) 3/4 a1F a = F / (m1+ m2)m1m2So a = (2/3)F / m1but F/m = aa = 2/3 a1Physics 111: Lecture 4, Pg #ForcesWe will consider two kinds of forces:Contact force:This is the most familiar kind.I push on the desk.The ground pushes on the chair...

Action at a distance:GravityElectricityPhysics 111: Lecture 4, Pg #Contact forces:Objects in contact exert forces.

Convention: Fa,b means the force acting on a due to b.

So Fhead,thumb means the force on the head due to the thumb.

Fhead,thumbPhysics 111: Lecture 4, Pg #Action at a distance

Gravity:Physics 111: Lecture 4, Pg #Gravitation(Courtesy of Newton)Newton found that amoon / g = 0.000278and noticed that RE2 / R2 = 0.000273

This inspired him to propose the Universal Law of Gravitation: |FMm |= GMm / R2RREamoongwhere G = 6.67 x 10 -11 m3 kg-1 s-2Physics 111: Lecture 4, Pg #Gravity...The magnitude of the gravitational force F12 exerted on an object having mass m1 by another object having mass m2 a distance R12 away is:

The direction of F12 is attractive, and lies along the line connecting the centers of the masses.

R12m1m2F12F21Physics 111: Lecture 4, Pg #Gravity...Near the Earths surface:R12 = REWont change much if we stay near the Earth's surface.i.e. since RE >> h, RE + h ~ RE.REmMh

FgPhysics 111: Lecture 4, Pg #Gravity...Near the Earths surface...

So |Fg| = mg = ma

a = g

All objects accelerate with acceleration g, regardless of their mass!

Where: =g

Leaky CupPhysics 111: Lecture 4, Pg #Example gravity problem:What is the force of gravity exerted by the earth on a typical physics student?

Typical student mass m = 55kgg = 9.8 m/s2.Fg = mg = (55 kg)x(9.8 m/s2 )

Fg = 539 N

FgThe force that gravity exerts on any object is called its Weight W = 539 NPhysics 111: Lecture 4, Pg #Lecture 4, Act 2Force and accelerationSuppose you are standing on a bathroom scale in 141 Loomis and it says that your weight is W. What will the same scale say your weight is on the surface of the mysterious Planet X ? You are told that RX ~ 20 REarth and MX ~ 300 MEarth.

(a) 0.75 W (b) 1.5 W

(c) 2.25 W EXPhysics 111: Lecture 4, Pg #Lecture 4, Act 2SolutionThe gravitational force on a person of mass m by another object (for instance a planet) having mass M is given by:

Ratio of weights = ratio of forces:

Physics 111: Lecture 4, Pg #Newtons Third Law:Forces occur in pairs: FA ,B = - FB ,A.

For every action there is an equal and opposite reaction.

We have already seen this in the case of gravity:

R12m1m2F12F21

NewtonsSailboardPhysics 111: Lecture 4, Pg #Newton's Third Law... FA ,B = - FB ,A. is true for contact forces as well:Fm,wFw,mFm,f Ff,m

2 SkateboardsPhysics 111: Lecture 4, Pg #Example of Bad ThinkingSince Fm,b = -Fb,m, why isnt Fnet = 0 and a = 0 ?a ??Fm,bFb,micePhysics 111: Lecture 4, Pg #Example of Good ThinkingConsider only the box as the system!Fon box = mabox = Fb,mFree Body Diagram (next time).aboxFm,bFb,micePhysics 111: Lecture 4, Pg #Lecture 4, Act 3Newtons 3rd LawTwo blocks are stacked on the ground. How many action-reaction pairs of forces are present in this system?(a) 2

(b) 3

(c) 4abPhysics 111: Lecture 4, Pg #Lecture 4, Act 3Solution:(c) 4FE,aFa,EabFE,babFb,EFb,aFa,babFg,bFb,gabPhysics 111: Lecture 4, Pg #Recap of todays lectureNewtons 3 Laws: (Text: 4-1 to 4-5)

Law 1: An object subject to no external forces is at rest or moves with a constant velocity if viewed from an inertial reference frame. (Text: 4-1)

Law 2: For any object, FNET = F = ma (Text: 4-2 & 4-3)

Law 3: Forces occur in pairs: FA ,B = - FB ,A. (Text: 4-4 & 4-5)

Look at Textbook problems Chapter 4: # 3, 5, 7, 19

ExtinguisherCartPhysics 111: Lecture 4, Pg #