Phys 207 Announcements z

Exam 2 this Friday, April 8th

Today’s Agenda z

z z

Review: ÍConservation of “total mechanical energy” ÍNon-conservative forces (friction) General work/energy theorem Problems using work/energy theorem

1

Conservation of Energy z

If only conservative forces are present, the total kinetic plus potential energy of a system is conserved, i.e. the total “mechanical energy” is conserved. Í(note: E=Emechanical throughout this discussion) E=K+U ∆E = ∆K + ∆U ⇒ using ∆K = W = W + ∆U = W + (-W) = 0 ⇒ using ∆U = -W

E = K + U is constant!!! z

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

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Page 1

Problem: Hotwheel z

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. Í Find v1 and v2.

v2 d h

v1

3

Problem: Hotwheel... z z z

K+U energy is conserved, so ∆E = 0 ∆K = - ∆U Moving down a distance d, ∆U = -mgd, ∆K = 1/2mv12 Solving for the speed: v1 = 2 gd

d

h

v1

4

Page 2

Problem: Hotwheel... z z z

At the end, we are a distance d - h below our starting point. ∆U = -mg(d - h), ∆K = 1/2mv22 Solving for the speed: v 2 = 2 g (d − h )

d-h

v2

d

h

5

Non-conservative Forces: Friction

z

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 = -µmgL. Clearly, the work done depends on the path taken.

z

Wpath 2 > Wpath 1

z

B path 1 path 2 A

z

When non-conservative forces act then energy can be dissipated into other modes (thermal,sound)

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Page 3

Generalized Work/Energy Theorem: z

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

z

The total work done is: WNET = WC + WNC

z

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

z

But WC = -∆U So

WNC = ∆K + ∆U = ∆Emechanical

7

Generalized Work/Energy Theorem: WNC = ∆K + ∆U = ∆Emechanical z

The change in kinetic+potential energy of a system is equal to the work done on it by non-conservative forces. Emechanical =K+U of system not conserved! ÍIf all the forces are conservative, we know that K+U energy is conserved: ∆K + ∆U = ∆Emechanical = 0 which says that WNC = 0, which makes sense. ÍIf some non-conservative force (like friction, a “push” or a “pull”) does work, K+U energy will not be conserved and WNC = ∆E, which also makes sense.

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Page 4

Problem: Block Sliding with Friction z

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 µk. Í How far, x, does the block go along the bottom portion of the track before stopping?

d

µk x 9

Problem: Block Sliding with Friction... z z z z z

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

x = d / µk d

µk

x 10

Page 5

More Example Problems z

Problems using work/energy theorem ÍSpring shot ÍEscape velocity ÍLoop the loop ÍVertical springs

z

Note: Lecture notes will have additional, related examples worked out.

11

Problem: Spring Shot z

A sling shot is made from a pair of springs each having spring constant k. The initial length of each spring is x0. A puck of mass m is placed at the point connecting the two springs and pulled back so that the length of each spring is x1. The puck is released. What is its speed v after leaving the springs? (The relaxed length of each spring is xr).

xr

x1

x0 m

m

m v

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Page 6

Problem: Spring Shot z

Only conservative forces are at work, so K+U energy is conserved. EI = EF ∆K = -∆Us

(

) (

1 2 2 2 2 ∆U s = 2 ⋅ k ( x0 − x r ) − ( x1 − x r ) = k ( x0 − x r ) − ( x1 − x r ) 2

)

x1 x0 m

m

13

Problem: Spring Shot z

Only conservative forces are at work, so K+U energy is ∆K = -∆Us conserved. EI = EF ∆K =

1 mv 2 2

m

m at rest

v

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Page 7

Problem: Spring Shot z

Only conservative forces are at work, so K+U energy is conserved. EI = EF ∆K = -∆Us

(

1 2 2 mv 2 = −k ( x0 − x r ) − ( x1 − x r ) 2

m

)

m v

15

Problem: How High? z

A projectile of mass m is launched straight up from the surface of the earth with initial speed v0. What is the maximum distance from the center of the earth RMAX it reaches before falling back down.

RMAX RE

m

v0

M 16

Page 8

Problem: How High... z

All forces are conservative: ÍWNC = 0 Í∆K = -∆U

RMAX RE

z

And we know: 1 ∆K = − mv 02 2

⎛ 1 1 − ∆U = −GMm⎜ − ⎜R R ⎝ E MAX

m

⎞ ⎟ ⎟ ⎠

v0 hMAX ⎛ 1 1 1 mv 02 = GMm ⎜⎜ − 2 ⎝ RE RMAX

M

⎞ ⎟ ⎟ ⎠

17

Problem: How High... ⎛ 1 1 1 − mv 02 = GMm⎜⎜ 2 ⎝ RE RMAX ⎛ 1 1 − v 02 = 2GM ⎜ ⎜R ⎝ E RMAX

⎞ ⎟ ⎟ ⎠

RMAX

⎞ ⎟ ⎟ ⎠

⎛ GM ⎞ ⎛ R = 2 ⎜ 2 ⎟RE ⎜1 − E ⎜R ⎟ ⎜ RMAX ⎝ E ⎠ ⎝ ⎛ R = 2 gRE ⎜1 − E ⎜ RMAX ⎝ v 02 R =1 − E 2 gRE RMAX

⎞ ⎟ ⎟ ⎠ RE

m

v0 hMAX

⎞ ⎟ ⎟ ⎠

M

R MAX =

RE v 02 1− 2 gRE 18

Page 9

Escape Velocity R MAX =

RE v 02 1− 2 gRE

If we want the projectile to escape to infinity we need to make the denominator in the above equation zero:

z

1−

v 02 =0 2 gR E

v 02 =1 2 gRE

v 0 = 2 gRE

We call this value of v0 the escape velocity, vesc

19

Escape Velocity z

GM we find the escape velocity from RE2 GM p a planet of mass Mp and radius Rp to be: v esc = 2 Rp (where G = 6.67 x 10-11 m3 kg-1 s-2). Remembering that g =

Rp(m)

Mp(kg)

gp(m/s2)

vesc(m/s)

Earth

6.378x106

5.976x1024

9.81

11.2x103

Moon

1.737x106

7.349x1022

1.62

2.38x103

Jupiter

7.149x107

1.900x1027

24.8

59.5x103

Sun

6.950x108

1.989x1030

275

618.x103 20

Page 10

Lecture 18, Act 1 Escape Velocity z

Two identical spaceships are awaiting launch on two planets with the same mass. Planet 1 is stationary, while Planet 2 is rotating with an angular velocity ω. ÍWhich spaceship needs more fuel to escape to infinity?

(a) 1

(b) 2

(c) same z

(1)

ω

(2) 21

Lecture 18, Act 1 Solution z z z z

Both spaceships require the same escape velocity to reach infinity. Thus, they require the same kinetic energy. Both initially have the same potential energy. Spaceship 2 already has some kinetic energy due to its rotational motion, so it requires less work (i.e. less fuel).

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Page 11

Lecture 18, Act 1 Algebraic Solution WNC = ∆K + ∆U = ∆E For spaceship 1:

W1 = (Kf - K0) + (Uf - U0)

K0 = 0 , U0 = 0

W1 = Kf + Uf

For spaceship 2:

W2 = (Kf - K0) + (Uf - U0) =

K0 =

1 m(ωr)2, U0 = 0 2

W2 = Kf + Uf -

1 m(ωr)2 2

W1 > W2 So spaceship 1 will need more fuel. 23

v = ωr

z

Lecture 18, Act 1 Aside

This is one of the reasons why all of the world’s spaceports are located as close to the equator as possible.

r1

r2 > r1

r2

K2 =

1 1 m(ωr2)2 > K1 = m(ωr1)2 2 2

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Page 12

Problem: Space Spring z

A low budget space program decides to launch a 10,000 kg spaceship into space using a big spring. If the spaceship is to reach a height RE above the surface of the Earth, what distance d must the launching spring be compressed if it has a spring constant of 108 N/m.

d

k

25

Problem: Space Spring... z z z

Since gravity is a conservative force, energy is conserved. Since K = 0 both initially and at the maximum height (v = 0) we know: Ubefore = Uafter (US + UG )before = (UG )after

⎛ 1 2 GMm ⎞ ⎛ GMm ⎞ ⎟ = ⎜− ⎟ ⎜ kd − ⎜ RE ⎟⎠ ⎝⎜ 2 RE ⎟⎠ ⎝2 1 GMm 2 GMm GMm kd 2 = − + = 2 2 RE 2 RE 2 RE

d =

GMm kRE

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Page 13

Problem: Space Spring So we find

d =

GMm kRE

z

For the numbers given, d = 79.1 m

z

But don’t get too happy...

F = kd = ma a = kd/m a = 79.1 x 106 m/s2 a = 791000 g unhappy astronaut!

d a

k

27

Problem: Loop the loop z

A mass m starts at rest on a frictionless track a distance H above the floor. It slides down to the level of the floor where it encounters a loop of radius R. What is H if the mass just barely makes it around the loop without losing contact with the track.

H R 28

Page 14

Problem: Loop the loop z z z z

Draw a FBD of the mass at the top of the loop: FTOT = -(mg+N) j ma = -mv2/R j If it “just” makes it, N = 0. Ímg = mv2/R

N

v mg

v = Rg v j

H R

i

29

v = Rg

Problem: Loop the loop z z z

Now notice that K+U energy is conserved. ∆K = -∆U. ∆U = -mg(h) = -mg(H-2R) ∆K = 1/2 mv2 = 1/2 mRg

H=

mg(H-2R) = 1/2 mRg

5 R 2

h = H - 2R v H R 30

Page 15

Lecture 18, Act 2 Energy Conservation z

A mass starts at rest on a frictionless track a distance H above the floor. It slides down to the level of the floor where it encounters a loop of radius R. What is H if the normal force on the block by the track at the top of the loop is equal to the weight of the block ?

(a) 3R

(b) 3.5R

(c) 4R

H R 31

Lecture 18, Act 2 Solution z z z z

Draw a FBD of the mass at the top of the loop: FNET = -(mg+N) j ma = -mv2/R j In this case, N = mg. Í2mg = mv2/R

N

v mg

v 2 = 2Rg

v j

H R

Page 16

i

32

v 2 = 2Rg

Lecture 18, Act 2 Solution

z

Use the fact that K+U energy is conserved: ∆K = -∆U.

z

∆U = -mg(h) = -mg(H - 2R), ∆K = 1/2 mv2 = mRg

z

mg(H - 2R) = mRg

H = 3R

h = H - 2R v H R 33

Vertical Springs z

z

A spring is hung vertically. Its relaxed position is at y = 0 (a). When a mass m is hung from its end, the new equilibrium position is ye (b).

(b)

(a)

j

k

Recall that the force of a spring is Fs = -kx. In case (b) Fs = mg and x = ye: -kye - mg = 0

y=0

(ye < 0)

m mg = -kye mg

y = ye -kye

(ok since ye is a negative number) 34

Page 17

Vertical Springs z

U =

(b)

(a)

The potential energy of the springmass system is:

j

1 ky 2 + mgy + C 2

k

but mg = -kye

y=0 1 U = ky 2 − ky e y + C 2

choose C to make U=0 at y = ye: 0=

1 2 2 ky e − ky e + C 2

C=

m

y = ye

1 2 ky e 2

35

Vertical Springs z

(b)

(a)

So:

j 1 1 U = ky 2 - ky e y + ky e 2 2 2 1 2 2 = k y + y e - 2y e y 2

(

k

)

y=0

which can be written:

m

1 2 U = k (y − y e ) 2

y = ye

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Page 18