physics 111: lecture 12 today's agenda

Physics 111: Lecture 12, Pg 1

Physics 111: Lecture 12

Today's Agenda

Problems using work/energy theorem Spring shotEscape velocity Loop the loopVertical springs

Definition of Power, with example


Problem: Spring Shot

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).

x0

m

x1

mv

m

xr



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

x0

x1

m m

2r1

2r0

2r1

2r0s xxxxkxxxxk

212U




K mv12

2

vm mat rest




vm m

2r1

2r0

2 xxxxkmv21

SpringShot


Problem: How High?

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 v0

m

M


Problem: How High...

All forces are conservative:WNC = 0 K = -U

RMAX

v0

m

hMAX

And we know:

20mv

21K

MAXE R1

R1GMmU

RE

M

MAXE

20 R

1R1GMmmv

21


MAX

E

E

20

MAX

EE

MAX

EE2

E

MAXE

20

MAXE

20

RR

1gR2v

RR

1gR2

RR

1RRGM2

R1

R1GM2v

R1

R1GMmmv

21

RMAX

RE v0

m

hMAX

M

R RvgR

MAXE

E

1

202

Problem: How High...


Escape Velocity

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

R RvgR

MAXE

E

1

202

12

002

vgRE

vgRE

02

21 v gRE0 2

We call this value of v0 the escape velocity, vesc


Escape Velocity

Remembering that we find the escape velocity from

a planet of mass Mp and radius Rp to be:

(where G = 6.67 x 10-11 m3 kg-1 s-2).

vGMResc

p

p 2

g GMRE

2

Moon

Earth

Sun

Jupiter

Rp(m) Mp(kg) gp(m/s2) vesc(m/s)

6.378x106 5.976x1024

1.737x106 7.349x1022

7.149x107 1.900x1027

6.950x108 1.989x1030

9.81

1.62

24.8

275

11.2x103

2.38x103

59.5x103

618.x103


Lecture 12, Act 1Escape Velocity

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 .Which spaceship needs more fuel to escape to infinity?

(a) 1 (b) 2 (c) same

(1) (2)


Lecture 12, Act 1Solution

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).


Lecture 12, Act 1Aside

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

r1

r2

r2 > r1

K2 = m(r2)2 > 21

K1 = m(r1)221

v = r


Problem: Space Spring

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


Problem: Space Spring...

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

EE

2

R2GMm

RGMmkd

21

12 2

22 2

2kd GMmR

GMmR

GMmRE E E

d GMmkRE


Problem: Space Spring

For the numbers given, d = 79.1 m

But don’t get too happy...

d GMmkRE

d

k

a

So we find

F = kd = maa = kd/ma = 79.1 x 106 m/s2

a = 791000 gunhappy astronaut!


Problem: Loop the loop

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.

HR



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.

mg = mv2/R

HR

v

mg

N

v

i

j

v Rg



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

HR

h = H - 2R

v

H R 52

v Rg

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

Loopthe

Loop


Lecture 12, Act 2Energy Conservation

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 ?

RH

(a) 3R (b) 3.5R (c) 4R



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.

2mg = mv2/R

R

v

mg

N

v

i

j

v 2Rg2

H



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

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

mg(H - 2R) = mRg

R

h = H - 2R

v

H 3R

H

v 2Rg2


Vertical Springs

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).

y = 0

y = ye

j

k

m

(a) (b)

mg -kye

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

-kye - mg = 0 (ye < 0)

(ok since ye is a negative number)

mg = -kye


Vertical Springs

The potential energy of the spring-mass system is:

y = 0

y = ye

j

k

m

(a) (b)

U ky mgy C 12

2

Cykyky21U e

2

but mg = -kye

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

Ckyky210 2

e2

e 2eky

21C


Vertical Springs

So:

y = 0

y = ye

j

k

m

(a) (b)

( )= - +

= + -

22

22

1 12 21 22

e e

e e

U ky ky y ky

k y y y y

which can be written:

2eyyk21U


Vertical Springs

So if we define a new y coordinate system such that y = 0 is at the equilibrium position, ( y = y - ye ) then we get the simple result:

y = 0

j

k

m

(a) (b)

2eyyk21U

2ky21U


Vertical Springs

If we choose y = 0 to be at the equilibrium position of the mass hanging on the spring, we can define the potential in the simple form.

Notice that g does not appear in this expression!!By choosing our coordinates and

constants cleverly, we can hide the effects of gravity.

y = 0

j

k

m

(a) (b)

U ky12

2


U of Spring

-60

-40

-20

0

20

40

60

80

100

120

140

160

-10 -8 -6 -4 -2 0 2 4 6 8 10

y

U US = 1/2ky2


U of Gravity

-60

-40

-20

0

20

40

60

80

100

120

140

160

-10 -8 -6 -4 -2 0 2 4 6 8 10

y

U

UG = mgy


U of Spring + Gravity

-60

-40

-20

0

20

40

60

80

100

120

140

160

-10 -8 -6 -4 -2 0 2 4 6 8 10

y

U

UG = mgy

US = 1/2ky2

UNET = UG + US

shift due to mgy term

0ye


U of Spring + Gravity

-60

-40

-20

0

20

40

60

80

100

120

140

160

-10 -8 -6 -4 -2 0 2 4 6 8 10

y

U

US = 1/2ky2

UNET =UG + US + C

0ye

shift due to mgy term

Choose C such as toshow that the newequilibrium positionhas zero potentialenergy:


Lecture 12, Act 3Energy Conservation

In (1) a mass is hanging from a spring. In (2) an identical mass is held at the height of the end of the same spring in its relaxed position. Which correctly describes the relation of the potential

energies of the two cases?

(a) U1 > U2 (b) U1 < U2 (c) U1 = U2

d

case 1 case 2



y = 0, U1 = 0

In case 1, it is simplest to choose the mass to have zero total potential energy (sum of spring and gravitational potential energies) at its equilibrium position.

The answer is (b) U1 < U2.

y = d d

relaxed

22 kd

21 U In case 2 the total potential energy is then .


Vertical Springs:Example Problem

If we displace the mass a distance d from equilibrium and let it go, it will oscillate up & down. Relate the maximum speed of the mass v to d and the spring constant k.

Since all forces are conservative,E = K + U is constant.

y = 0

j

k

m y = d

y = -d

vU ky

12

2 2mv21K We know:


Vertical Springs:Example Problem

At the initial stretched positionand K = 0 (since v=0).

y = 0

j

k

m

= 212

E kd Since E=K+U is conserved,will always be true !

y = d

y = -d

v

U kd12

2

Energy is shared between the K and U terms. At y = d or -d the energy is all potential At y = 0, the energy is all kinetic.

12

12

2 2kd mv v d km

222 mv21ky

21kd

21E

Spring


Power

We have seen that W = F.rThis does not depend on time!

Fr

v

P dWdt

Power is the “rate of doing work”:

If the force does not depend on

time: dW/dt = F.dr/dt = F.v P = F.v Units of power: J/sec = N-m/sec = Watts

Ladder


Power A 2000 kg trolley is pulled up

a 30 degree hill at 20 mi/hrby a winch at the top of thehill. How much power is thewinch providing?

The power is P = F.v = T.v

Since the trolley is not accelerating, the net force on it must be zero. In the x direction:T - mg sin = 0T = mg sin

v

mg

Twinch

xy


Power

P = T.v = Tv since T is parallel to v

So P = mgv sin

v = 20 mi/hr = 8.94 m/sg = 9.81 m/s2

m = 2000 kgsin = sin(30o) = 0.5

and P = (2000 kg)(9.81 m/s2)(8.94 m/s)(0.5) = 87,700 W

v

mg

Twinch

xy


Recap of today’s lecture

Problems using work/energy theoremSpring shotEscape velocity Loop the loopVertical springs

Definition of Power, with example (Text: 6-3)

Look at textbook problems Chapter 8: # 81, 85, 95 Chapter 11: # 49

physics 111: lecture 12 today's agenda

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