physics 1501: lecture 14, pg 1 physics 1501: lecture 14 today’s agenda l midterm graded by next...

Physics 1501: Lecture 14, Pg 1

Physics 1501: Lecture 14Physics 1501: Lecture 14TodayToday’’s Agendas Agenda

Midterm graded by next Monday (maybe …)

Homework #5: Due Friday Oct. 7 @ 11:00 AM

TopicsConservative vs non-conservative forcesConservation of mechanical energythe U - F relationship


Some DefinitionsSome Definitions

Conservative Forces - those forces for which the work done does not depend on the path taken, but only the initial and final position.

Potential Energy - describes the amount of work that can potentially be done by one object on another under the influence of a conservative force

W = -U

only differences in potential energy matter.


Lecture 14,Lecture 14, ACT 1ACT 1Work/Energy for Non-Conservative ForcesWork/Energy for Non-Conservative Forces

The air track is once again at an angle of 30 degrees with respect to horizontal. The cart (with mass 1 kg) is released 1 meter from the bottom and hits the bumper with some speed, v1. This time the vacuum/ air generator breaks half-way through and the air stops. The cart only bounces up half as high as where it started. How much work did friction do on the cart ?

1 meter

30 degrees

A) 2.5 J B) 5 J C) 10 J D) –2.5 J E) –5 J F) –10 J


Conservation of EnergyConservation of Energy If only conservative forces are present, the total If only conservative forces are present, the total

energy (sum of potential and kinetic energies) of energy (sum of potential and kinetic energies) of a systema system is conserved.is conserved. E = K + U

E = K + U = W + U = W + (-W) = 0

using K = Wusing U = -W

Both K and U can change, but E = K + U remains constant.

E = K + U is constantconstant !!!!!!


Example: The simple pendulum.Example: The simple pendulum.

Suppose we release a mass m from rest a distance h1 above its lowest possible point. What is the maximum speed of the mass and where

does this happen ? To what height h2 does it rise on the other side ?

v

h1 h2

m



Energy is conserved since gravity is a conservative force (E = K + U is constant)

Choose y = 0 at the bottom of the swing, and U = 0 at y = 0 (arbitrary choice).

E = 1/2mv2 + mgy.

v

h1 h2

y

y=0


Example: The simple pendulum.Example: The simple pendulum. E = 1/2mv2 + mgy.

Initially, y = h1 and v = 0, so E = mgh1. Since E = mgh1 initially, E = mgh1 always since energy is conserved.

y

y=0

y=h1



1/2mv2 will be maximum at the bottom of the swing. So at y = 0 1/2mv2 = mgh1 v2 = 2gh1

v

h1

y

y=h1

y=0



Since E = mgh1 = 1/2mv2 + mgy it is clear that the maximum

height on the other side will be at y = h1=h2 and v = 0. The ball returns to its original height.

y

y=h1=h2

y=0



The ball will oscillate back and forth. The limits on its height and speed are a consequence of the sharing of energy between K and U.

E = 1/2mv2 + mgy = K + U = constant.

y

Active Figure


What speed will skateboarder reach at bottom of the hill ?

R=3 m

. .

m = 25 kg

Initial: K1 = 0 U1 = mgR

Final: K2 = 1/2 mv2 U2 = 0

Conservation ofTotal Energy :

Lecture 14, Lecture 14, ExampleExampleSkateboardSkateboard

K1 + U1 = K2 + U2

0 + mgR = 1/2mv2 + 0

v = (2gR)1/2

v ~ (2 x10m/s2 x 3m)1/2

v ~ 8 m/s (~16mph) !

Does NOT depend on the mass !

. .


What would be the speed if instead the skateboarder jumps to the ground on the other side ?

. .

. .

R=3 m

KINEMATICS:

* v = a t * h = 1/2 at2

2h = v2/a v = (2 h a)1/2 = (2 R g)1/2

the same magnitude as before !

and independent of mass



Lecture 14, Lecture 14, ACT 2ACT 2The Roller CoasterThe Roller Coaster

I have built a Roller Coaster . A motor tugs the cars to the top and then they are let go and are in the hands of gravity. To make the following loop, how high do I have to let the release the car ??

h ?

R

Car has mass mHint: from lecture 11

we know that to avoid death, the minimum speed at the top is

A) 2R B) 5R C) 5/2 R D) none of the above


Non-conservative Forces :Non-conservative Forces :

If the work done does not depend on the path taken, the force involved is said to be conservative.

If the work done does depend on the path taken, the force involved is said to be non-conservative.

An example of a non-conservative force is friction:

Pushing a box across the floor, the amount of work that is done by friction depends on the path taken. Work done is proportional to the length of the path !


Generalized Work Energy Theorem:Generalized Work Energy Theorem:

Suppose FNET = FC + FNC (sum of conservative and non-conservative forces).

The total work done is: WTOT = WC + WNC

The Work Kinetic-Energy theorem says that: WTOT = K. WTOT = WC + WNC = K WNC = K - WC

But WC = -U

So WNC = K + U = E


Lecture 14, Lecture 14, ACT 3ACT 3Stones and FrictionStones and Friction

I throw a stone into the air. While in flight it feels the force of gravity and the frictional force of the air resistance. The time the stone takes to reach the top of its flight path (i.e. go up) is,

A) larger than

B) equal to

C) less than

The time it takes to return from the top (i.e. go down).


Let’s now suppose that the surface is not frictionless and the same skateboarder reach the speed of 7.0 m/s at bottom of the hill. What was the work done by friction on the skateboarder ?

R=3 m

. .

m = 25 kg

Conservation ofTotal Energy :


K1 + U1 = K2 + U2

Wf + 0 + mgR = 1/2mv2 + 0

Wf = 1/2mv 2 - mgR

Wf = (1/2 x25 kg x (7.0 m/s2)2 - - 25 kg x 10m/s2 3 m)

Wf = 613 - 735 J = - 122 J

Total mechanical energy decreased by 122 J !

. .

Wf +


Conservative Forces Conservative Forces and Potential Energyand Potential Energy

We have defined potential energy for conservative forcesU = -W

But we also now that

W = Fxx

Combining these two,

U = - Fxx

Letting small quantities go to infinitessimals,

dU = - Fxdx

Or,

Fx = -dU/dx


Examples of the U - F relationshipExamples of the U - F relationship

Remember the spring,U = (1/2)kx2

Do the derivative

Fx = - d ( (1/2)kx2) / dx

Fx = - 2 (1/2) kx

Fx = -kx


Examples of the U - F relationshipExamples of the U - F relationship

Remember gravity,

Fy = mg

Do the integral

U = - F dyU = - (-mg) dyU = mgy + CU = U2 – U1 = (mgy2 + C) – (mgy1 + C) = mgy

physics 1501: lecture 14, pg 1 physics 1501: lecture 14 today’s agenda l midterm graded by next...

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