physics 111: lecture 11, pg 1 physics 111: lecture 11 today’s agenda l review l work done by...

Physics 111: Lecture 11, Pg 1

Physics 111: Lecture 11

Today’s Agenda

Review Work done by variable force in 3-D

Newton’s gravitational force Conservative forces & potential energy Conservation of “total mechanical energy”

Example: pendulum Non-conservative forces

friction General work/energy theorem Example problem


Work by variable force in 3-D:

Work dWF of a force F acting

through an infinitesimaldisplacement r is:

dW = F.r The work of a big displacement through a variable force will

be the integral of a set of infinitesimal displacements:

WTOT = F.r

F

r

ò


Work by variable force in 3-D:Newton’s Gravitational Force

Work dWg done on an object by gravity in a displacement dr isgiven by:

dWg = Fg.dr = (-GMm / R2 r).(dR r + Rd)

dWg = (-GMm / R2) dR (since r. = 0, r.r = 1)

^ ^^

r^

^drRd

dR

R

Fg m

M

d

^ ^ ^ ^



Integrate dWg to find the total work done by gravity in a “big”displacement: Wg = dWg = (-GMm / R2) dR = GMm (1/R2 - 1/R1)

Fg(R1)

R1

R2

Fg(R2)

R1

R2

R1

R2

m

M



Work done depends only on R1 and R2, not on the path taken.

R1

R2

m

M

12

g R

1

R

1GMmW


Lecture 11, Act 1Work & Energy

A rock is dropped from a distance RE above the surface of the earth, and is observed to have kinetic energy K1 when it hits the ground. An identical rock is dropped from twice the height (2RE) above the earth’s surface and has kinetic energy K2 when it hits. RE is the radius of the earth.

What is K2 / K1?

(a)

(b)

(c)

32

43

2

RE

RE

2RE


Lecture 11, Act 1Solution

Since energy is conserved, DK = WG.

RE

RE

2RE

12G R

1R1

GMm=W

12 R1

R1

c=ΔK

Where c = GMm is the same for both rocks


Lecture 11, Act 1 Solution

For the first rock:

RE

RE

EEE1 R

121

c2R

1R1

c=K

12 R1

R1

c=ΔK

2RE

EEE2 R

132

c3R

1R1

c=K

For the second rock:

So:34

2132

=KK

1

2


Newton’s Gravitational ForceNear the Earth’s Surface:

Suppose R1 = RE and R2 = RE + y

but we have learned that

So: Wg = -mgy

y

R

GMm

RyR

RyRGMm

RR

RRGMmW

2

EEE

EE

21

12g

GM

Rg

E2

RE+ y

M

m

RE


Conservative Forces:

We have seen that the work done by gravity does not depend on the path taken.

R1

R2

M

m

hm

Wg = -mgh

12

g R

1

R

1GMmW


Conservative Forces:

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

Gravity is a conservative force:

Gravity near the Earth’s surface:

A spring produces a conservative force: 2

1

2

2sxxk

2

1W

ymgWg

12

g R

1

R

1GMmW


Conservative Forces: We have seen that the work done by a conservative force

does not depend on the path taken.

W1

The work done can be “reclaimed.”

W2

W1

W2

W1 = W2

WNET = W1 - W2 = 0

Therefore the work done in a closed path is 0.


Lecture 11, Act 2Conservative Forces

The pictures below show force vectors at different points in space for two forces. Which one is conservative ?

(a) 1 (b) 2 (c) both

(1) (2)

y

x

y

x



Consider the work done by force when moving along different paths in each case:

(1) (2)

WA = WB WA > WB


Lecture 11, Act 2 In fact, you could make money on type (2) if it ever existed:

Work done by this force in a “round trip” is > 0!Free kinetic energy!!

W = 15 J

W = 0

W = -5 J

W = 0WNET = 10 J = DK


Potential Energy

For any conservative force F we can define a potential energy function U in the following way:

The work done by a conservative force is equal and opposite to the change in the potential energy function.

This can be written as:

W = F.dr = -U ò

U = U2 - U1 = -W = - F.dròr1

r2

r1

r2 U2

U1


Gravitational Potential Energy

We have seen that the work done by gravity near the Earth’s surface when an object of mass m is lifted a distance y is Wg = -mg y

The change in potential energy of this object is therefore:

U = -Wg = mg y

ym

Wg = -mg y j


Gravitational Potential Energy

So we see that the change in U near the Earth’s surface is: U = -Wg = mg y = mg(y2 -y1).

So U = mg y + U0 where U0 is an arbitrary constant.

Having an arbitrary constant U0 is equivalent to saying that we can choose the y location where U = 0 to be anywhere we want to.

y1

m

Wg = -mg y j y2


Potential Energy Recap:

For any conservative force we can define a potential energy function U such that:

The potential energy function U is always defined onlyup to an additive constant.

You can choose the location where U = 0 to be anywhere convenient.

U = U2 - U1 = -W = - F.drS1

S2


Conservative Forces & Potential Energies (stuff you should know):

Force F

WorkW(1-2)

Change in P.E

U = U2 - U1

P.E. functionU

Fg = -mg j

Fg = r

Fs = -kx

^

^

12R

1

R

1GMm

2

1

2

2xxk

2

1

-mg(y2-y1) mg(y2-y1)

12R

1

R

1GMm

2

1

2

2xxk

2

1

2R

GMm

mgy + C

GMm

RC

1

22kx C


Lecture 11, Act 3Potential Energy

All springs and masses are identical. (Gravity acts down). Which of the systems below has the most potential energy

stored in its spring(s), relative to the relaxed position?

(a) 1

(b) 2

(c) same

(1) (2)



The displacement of (1) from equilibrium will be half of that of (2) (each spring exerts half of the force needed to balance mg)

(1) (2)

d 2d

0



(1) (2)

d 2d

0

The potential energy stored in (1) is 212

kd kd2 2

The potential energy stored in (2) is 12

k 2d 2kd2 2

The spring P.E. is twice as big in (2) !


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

plus potential energy of a system is conserved.

E = K + U is constant!!!

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

E = K + U

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

using K = Wusing U = -W


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



Kinetic+potential 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



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



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

v gh 2 1

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

Bowling



We can also solve this by choosing y = 0 to be at the original position of the mass, and U = 0 at y = 0.

E = 1/2mv2 + mgy.

v

h1 h2

y

y = 0



E = 1/2mv2 + mgy. Initially, y = 0 and v = 0, so E = 0. Since E = 0 initially, E = 0 always since energy is conserved.

y

y = 0



1/2mv2 will be maximum at the bottom of the swing. So at y = -h1 1/2mv2 = mgh1 v2 = 2gh1

v

h1

y

y = 0

y = -h1

v gh 2 1

Same as before!



Since 1/2mv2 - mgh = 0 it is clear that the maximum height on the other side will be at y = 0 and v = 0.

The ball returns to its original height.

y

y = 0

Same as before!

Galileo’sPendulum


Example: Airtrack & Glider

A glider of mass M is initially at rest on a horizontal frictionless track. A mass m is attached to it with a massless string hung over a massless pulley as shown. What is the speed v of M after m has fallen a distance d ?

d

M

m

v

v


Example: Airtrack & Glider

Kinetic+potential energy is conserved since all forces are conservative.

Choose initial configuration to have U=0.

K = -U

d

M

m

v

1

22m M v mgd

Glider


Problem: Hotwheel

A toy car slides on the frictionless track shown below. It starts at rest, drops a distance d, moves horizontally at speed v1, rises a distance h, and ends up moving horizontally with speed v2. Find v1 and v2.

hd

v1

v2


Problem: Hotwheel...

K+U energy is conserved, so E = 0 K = - U Moving down a distance d, U = -mgd, K = 1/2mv1

2

Solving for the speed:

hd

v1

v gd1 2


Problem: Hotwheel...

At the end, we are a distance d - h below our starting point. U = -mg(d - h), K = 1/2mv2

2

Solving for the speed:

hd

v2

v g d h2 2

d - h


Non-conservative Forces:

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

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

An example of a non-conservative force is friction.When 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!


Non-conservative Forces: Friction

Suppose you are pushing a box across a flat floor. The mass of the box is m and the coefficient of kinetic friction is k.

The work done in pushing it a distance D is given by:

Wf = Ff • D = -kmgD.

D

Ff = -kmg


Non-conservative Forces: Friction

Since the force is constant in magnitude and opposite in direction to the displacement, the work done in pushing the box through an arbitrary path of length L is just Wf = -mgL.

Clearly, the work done depends on the path taken.

Wpath 2 > Wpath 1

A

B

path 1

path 2


Generalized Work/Energy Theorem:

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

The total work done is: WNET = WC + WNC

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

But WC = -U

So WNC = K + U = E


Generalized Work/Energy Theorem:

The change in kinetic+potential energy of a system is equal to the work done on it by non-conservative forces. E=K+U of system not conserved!

If all the forces are conservative, we know that K+U energy is conserved: K + U = E = 0 which says that WNC = 0, which makes sense.

If some non-conservative force (like friction) does work,K+U energy will not be conserved and WNC = E, which also makes sense.

WNC = K + U = E


Problem: Block Sliding with Friction

A block slides down a frictionless ramp. Suppose the horizontal (bottom) portion of the track is rough, such that the coefficient of kinetic friction between the block and the track is k. How far, x, does the block go along the bottom portion

of the track before stopping?

x

d k


Problem: Block Sliding with Friction...

Using WNC = K + U As before, U = -mgd WNC = work done by friction = -kmgx. K = 0 since the block starts out and ends up at rest. WNC = U -kmgx = -mgd

x = d / k

x

d k


Recap of today’s lecture

Work done by variable force in 3-D (Text: 6-1) Newton’s gravitational force (Text: 11-2)

Conservative Forces & Potential energy (Text: 6-4)

Conservation of “Total Mechanical Energy” (Text: 7-1) Examples: pendulum, airtrack, Hotwheel car

Non-conservative forces (Text: 6-4) friction

General work/energy theorem (Text: 7-2)

Example problem

Look at Textbook problems Chapter 7: # 9, 21, 27, 51, 77

physics 111: lecture 11, pg 1 physics 111: lecture 11 today’s agenda l review l work done by...

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