Physics 2210 Fall 2015

smartPhysics 08 Conservative Force and Potential Energy

09 Work and Potential Energy, Part II 09/30/2015

Conservative forces Definition: Forces whose work done an object is (always) path-independent are called conservative forces The work done by a conservative force from point 1 to

any other point 2, and back to 1 along any closed loop is ALWAYS ZERO.

i.e. for any conservative force �⃗�𝐶

� �⃗�𝐶 ∙ 𝑑𝑟 ≡ 0

Integrating over a closed loop

Identically zero: True for any loop

Subscript C indicates a conservative force

Example of non-conservative forces 1. Forces (of magnitude 𝐹𝐴) exerted by a hand in moving an object at

constant speed on a rough surface through a closed loop 2. The kinetic force of friction (of magnitude 𝑓𝑠) on that same object

The applied force 𝑭𝑨 is always in the direction of motion: 𝑾𝑨 > 𝟎 over closed loop

The friction force 𝒇𝒌 is always opposite the direction of motion: 𝑾𝒇𝒌 < 𝟎 over closed loop

Unit 07

NOTE: 𝒙 is not spring length

This last point is confusing… don’t use it and don’t think about it. I prefer you remember the definition this way:

𝑈 𝑟 = 𝑈 𝑟0 −𝑊𝑟0→𝑟 = −𝑊𝑟0→𝑟 Example: gravity (CHOOSING 𝑈 = 0 at 𝑦 = 0)

𝑈 𝑦 = 𝑈 0 −𝑊0→𝑦 = 0 − −𝑚𝑚 𝑦 − 0 = 𝑚𝑚𝑦

Unit 08

Unit 08

The work done by the force of gravity (near surface of Earth), a uniformly constant force, is given by (assuming UP to be the +𝑦 direction)

𝑊𝑔 = �⃗�𝑔 ∙ ∆𝑟 𝑊𝑔 = −𝑚𝑚 ∆𝑦

𝑚

𝑓𝑘

𝑚𝑚

𝑁 𝑟𝑖

𝑟𝑓 𝑓𝑘

𝑚𝑚

𝑁

𝑓𝑘

𝑚𝑚

𝑁

𝑓𝑘

𝑚𝑚

𝑁

𝑟𝑖

𝑟𝑓

Vertical component of force of gravity �⃗�𝑔

Vertical component of displacement ∆𝑟

∆𝑦<0 here

Gravity is a conservative force

Potential Energy of gravity 𝑈𝑔 𝑦 − 𝑈𝑔 0 = −𝑊𝑔 = − −𝑚𝑚 ∆𝑦

𝑈𝑔 𝑦 = 𝑚𝑚𝑦 CHOOSE 𝑈𝑔 = 0 at 𝑦 = 0 and UP to be +𝑦

Poll 09-28-02

Three balls of equal mass are fired simultaneously with equal speeds from the same height h above the ground. Ball 1 is fired straight up, ball 2 is fired straight down, and ball 3 is fired horizontally. Rank the speeds of the balls, v1, v2, and v3, just before each ball hits the ground.

A. v2 > v3 > v1 B. v3 > v2 > v1 C. v1 > v2 > v3 D. v1 = v2 = v3

Example 08-01 (1/3) (a) The change in your gravitational potential energy on taking an

elevator from the ground floor to the top of the Empire State Building. The building is 102 stories high (assuming a 3 min ride to the top of the building). (Assuming your mass is 71 kg and the height of one story to be 3.5 m.) Give your answer in kJ

(b) Find the average force exerted by the elevator on you during the trip in newtons (N)

(c) Find the average power delivered by that force in kilowatts (kW)

(%i1) DPEg: m*g*H; (%o1) g m H

(%i2) DPEg, m=71, g=9.81, H=102*3.5;

(%o2) 248654.07

(%i3) /* change to kJ by multiplying 1 = 1kJ/1000J */

%/1000;

(%o3) 248.65407

Answer (a) 249 kJ

Aside: Power

Standard Horse Power: 1 hp = 745.7 W

Example 08-01 (3/3) (b) Find the average force exerted by the elevator on you during the trip. (c) Find the average power delivered by that force in kilowatts (kW)

(%i4) /* assuing constant speed then the force exerted by the elevator on you is equal to m*g upward to cancel your weight */

F: m*g;

(%o4) g m

(%i5) F, m=71, g=9.81;

(%o5) 696.51

Answer (b) 697 N

(%i6) /* Work done by elevator = F*H */

W: F*H;

(%o6) g m H

(%i7) /* power: P=W/t */

P: W/t;

g m H

(%o7) -----

t

(%i8) /* t=3 min=180s */

P, m=71, g=9.81, H=102*3.5, t=180;

(%o8) 1381.4115

(%i9) /* convert to kW by mupltiplying 1=1kW/1000 W */

%/1000;

(%o9) 1.3814115

Answer (c) 1.38 kW

Other Conservative Forces • All central forces are conservative • A central force is:

– one that is directed always towards or away from a “center” (we usually assign this to be the origin)

– whose magnitude does not depend on the orientation – Whose magnitude depends only on the distance to the

“center”

• Examples – Force exerted by a spring on a body tied to it at one end

(often the other end is tied to the Earth) – Force of gravity on a small object by another (often much

larger) object. – Electrical (actually: electrostatic) forces between charged

objects (PHYS 2220)

Frictionless, horizontal surface

𝑂

Spring of force constant 𝑘

relaxed length 𝐿0 𝑚

𝑟𝑖

𝑟𝑓

Path from 𝑟𝑖 to 𝑟𝑓

Approximate path by 2N that alternate 𝑛 = 1,2,3, …𝑁, (2𝑛 − 1)th: radial (2𝑛)th: arc

�⃗�𝑠 ⊥ ∆𝑟 for arcs: work done along the even segments vanish

�⃗� 𝑟2𝑛 ∙ ∆𝑟2𝑛= 0

�⃗�𝑠 ∥ ∆𝑟 for radial steps: Only work done along the odd segments contribute �⃗� 𝑟2𝑛−1 ∙ ∆𝑟2𝑛−1=

𝐹 𝑟2𝑛−1 ∙ ∆𝑟2𝑛−1

lim𝑁→∞

� 𝐹(𝑟)𝑑𝑟

𝑟𝑓

𝑟𝑖

≡ �𝐹 𝑟2𝑛−1 ∙ ∆𝑟2𝑛−1

𝑁

𝑛=1

𝑊𝑠 =

In the limit 𝑵 → ∞ the blue path becomes the black path

The green path gives the same work as the blue path

The red path also gives the same work as the green and blue paths

Top View

𝐹 𝑟 = −𝑘(𝑟 − 𝐿0)

Work done by a Spring 𝑊𝑠 ≡ � 𝐹𝑠 𝑟 𝑑𝑟

𝑟𝑓

𝑟𝑖

= � −𝑘 𝑟 − 𝐿0 𝑑𝑟

𝑟𝑓

𝑟𝑖

= −𝑘 � 𝑟 − 𝐿0 𝑑𝑟

𝑟𝑓

𝑟𝑖

Change of Variable 𝑥 = 𝑟 − 𝐿0, 𝑥 = 𝑑𝑟

𝑊𝑠 ≡ −𝑘 � 𝑥𝑑𝑥

𝑟𝑓−𝐿0

𝑟𝑖−𝐿0

= −12𝑘𝑥

2𝑟𝑖−𝐿0

𝑟𝑓−𝐿0

→ 𝑊𝑠 = −12𝑘 𝑟𝑓 − 𝐿02 − 𝑟𝑖 − 𝐿0 2

Or of we let 𝒙 represent the deformation (which we already did), then

𝑊𝑠 = −12𝑘 ∆𝐿𝑓

2 − ∆𝐿𝑖 2 , or 𝑊𝑠 = −12𝑘 𝑥𝑓2 − 𝑥𝑖2

𝑈𝑆 𝑟 − 𝑈𝑆 𝐿0 = −𝑊𝑠 𝐿0→𝑟 = − − 12𝑘 𝑟 − 𝐿0 2 − 𝐿0 − 𝐿0 2

= + 12𝑘 𝑟 − 𝐿0 2 − 0 2 = 12𝑘 𝑟 − 𝐿0 2

Taking the relaxed spring to have ZERO potential energy (usually best choice)

𝑈𝑆 = 12𝑘 ∆𝐿 2, or 𝑈𝑆 = 1

2𝑘𝑥2

Potential Energy of a Spring

NOTE: 𝒙 is not spring length

NOTE: 𝒙 is not spring length

Poll 09-30-01

A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed position while momentarily coming to rest.

If the initial speed of the box were doubled, how far x2 would the spring compress?

A. x2 = 2 x1 B. x2 = 2 x1 C. x2 = 4 x1

NOTE: 𝒙 is not spring length

Example 08-02 (1/3) A block of mass 𝑚 is pushed up against a spring, compressing it a distance 𝑥, and the block is then released. The spring projects the block along a frictionless horizontal surface, giving the block a speed 𝑣. The same spring is then used to project a second block of mass 4𝑚, giving it a speed of 5𝑣. What distance 𝑥2 was the spring compressed in the second case? Answer in terms of a numerical factor times 𝑥, the compression of the first block.

(%i1) /* total energy */

Energy: 0.5*mass*speed^2 + 0.5*k*compression^2;

2 2

(%o1) 0.5 mass speed + 0.5 compression k

(%i2) E1i: Energy, mass=m, speed=0, compression=x;

2

(%o2) 0.5 k x

(%i3) E1f: Energy, mass=m, speed=v, compression=0;

2

(%o3) 0.5 m v

(%i4) eqn1: E1i = E1f;

2 2

(%o4) 0.5 k x = 0.5 m v

(%i5) E2i: Energy, mass=4*m, speed=0, compression=x2;

2

(%o5) 0.5 k x2 ... continued

Example 08-02 (2/3) Spring on mass 𝑚 , compression 𝑥, results in speed 𝑣. same spring on mass 4𝑚: speed of 5𝑣. What distance was the spring compressed in the second case? i.e. 𝑥2/ 𝑥 = ?

(%i6) E2f: Energy, mass=4*m, speed=5*v, compression=0; 2

(%o6) 50.0 m v

(%i7) eqn2: E2i = E2f;

2 2

(%o7) 0.5 k x2 = 50.0 m v

(%i8) /* strategy: solve for k in each case */

soln1: solve(eqn1, k);

2

m v

(%o8) [k = ----]

2

x

(%i9) k1: rhs(soln1[1]), numer;

2

m v

(%o9) ----

2

x ... continued

Example 08-02 (3/3) Spring on mass 𝑚 , compression 𝑥, results in speed 𝑣. same spring on mass 4𝑚: speed of 5𝑣. What distance was the spring compressed in the second case? i.e. 𝑥2/ 𝑥 = ?

(%i10) soln2: solve(eqn2, k); 2

100 m v

(%o10) [k = --------]

2

x2

(%i11) k2: rhs(soln2[1]), numer;

2

100 m v

(%o11) --------

2

x2

(%i12) soln3: solve(k1=k2, x2);

(%o12) [x2 = - 10 x, x2 = 10 x]

(%i13) /* take positive root */

x2: rhs(soln3[2]);

(%o13) 10 x

Answer: 10

Material covered in homework for Unit 8 stops here

Universal Gravitation (in HW for unit 9)

𝑊𝐺 = � 𝐹𝑟 𝑟 𝑑𝑟

𝑟𝑓

𝑟𝑖

= � −𝐺𝐺𝑚𝑟2

𝑑𝑟

𝑟𝑓

𝑟𝑖

= −𝐺𝐺𝑚 � 𝑟−2𝑑𝑟

𝑟𝑓

𝑟𝑖

= −𝐺𝐺𝑚−1

𝑟−1 𝑟𝑖𝑟𝑓

𝑊𝐺 = 𝐺𝐺𝑚1𝑟𝑓−

1𝑟𝑖

FG

M

m 𝑟

We treat the gravitational force exerted by a very large spherical mass M on a small mass m as if M is stationary with its center at the origin. Then the force on m always points towards the origin, and with a radial component of

𝐹𝑟 = −𝐺𝐺𝑚𝑟2

The minus sign means it points inward

It is conventional to choose 𝑈G = 0 at 𝑟 = ∞

𝑈G = 0 − 𝐺𝐺𝑚1𝑟−

1∞

=−𝐺𝐺𝑚

𝑟

𝑈G =−𝐺𝐺𝑚

𝑟

Poll 09-30-02 (checkpoint for unit 8)

Consider two identical objects released from rest high above the surface of the earth (neglect air resistance for this question).

Case 1: we release an object from a height above the surface of the earth equal to 1 earth radius, its kinetic energy just before it hits the earth is K1.

Case 2 we release an object from a height above the surface of the earth equal to 2 earth radii, its kinetic energy just before it hits the earth to be K2.

Compare the kinetic energy of the two just before they hit the surface of the earth.

A. K2 = 2 K1 B. K2 = 4 K1 C. K2 = (4/3) K1 D. K2 = (3/2) K1

Unit 09

In your instructor’ terms:

∆𝐸 = � work done by non − conservative force 𝑗𝑗

Unit 09