physics 2210 fall 2015woolf/2210_jui/sept30.pdfto 1 earth radius, its kinetic energy just before it...
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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 ๏ฟฝโ๏ฟฝ๐ถ
๏ฟฝ ๏ฟฝโ๏ฟฝ๐ถ โ ๐๐ โก 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 ๐น๐ด) 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 ๐๐ ) on that same object
The applied force ๐ญ๐จ is always in the direction of motion: ๐พ๐จ > ๐ over closed loop
The friction force ๐๐ is always opposite the direction of motion: ๐พ๐๐ < ๐ over closed loop
Unit 07
NOTE: ๐ is not spring length
This last point is confusingโฆ donโt use it and donโt think about it. I prefer you remember the definition this way:
๐ ๐ = ๐ ๐0 โ๐๐0โ๐ = โ๐๐0โ๐ Example: gravity (CHOOSING ๐ = 0 at ๐ฆ = 0)
๐ ๐ฆ = ๐ 0 โ๐0โ๐ฆ = 0 โ โ๐๐ ๐ฆ โ 0 = ๐๐๐ฆ
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 +๐ฆ direction)
๐๐ = ๏ฟฝโ๏ฟฝ๐ โ โ๐ ๐๐ = โ๐๐ โ๐ฆ
๐
๐๐
๐๐
๐ ๐๐
๐๐ ๐๐
๐๐
๐
๐๐
๐๐
๐
๐๐
๐๐
๐
๐๐
๐๐
Vertical component of force of gravity ๏ฟฝโ๏ฟฝ๐
Vertical component of displacement โ๐
โ๐ฆ<0 here
Gravity is a conservative force
Potential Energy of gravity ๐๐ ๐ฆ โ ๐๐ 0 = โ๐๐ = โ โ๐๐ โ๐ฆ
๐๐ ๐ฆ = ๐๐๐ฆ CHOOSE ๐๐ = 0 at ๐ฆ = 0 and UP to be +๐ฆ
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 โข All central forces are conservative โข A central force is:
โ one that is directed always towards or away from a โcenterโ (we usually assign this to be the origin)
โ whose magnitude does not depend on the orientation โ Whose magnitude depends only on the distance to the
โcenterโ
โข Examples โ Force exerted by a spring on a body tied to it at one end
(often the other end is tied to the Earth) โ Force of gravity on a small object by another (often much
larger) object. โ Electrical (actually: electrostatic) forces between charged
objects (PHYS 2220)
Frictionless, horizontal surface
๐
Spring of force constant ๐
relaxed length ๐ฟ0 ๐
๐๐
๐๐
Path from ๐๐ to ๐๐
Approximate path by 2N that alternate ๐ = 1,2,3, โฆ๐, (2๐ โ 1)th: radial (2๐)th: arc
๏ฟฝโ๏ฟฝ๐ โฅ โ๐ for arcs: work done along the even segments vanish
๏ฟฝโ๏ฟฝ ๐2๐ โ โ๐2๐= 0
๏ฟฝโ๏ฟฝ๐ โฅ โ๐ for radial steps: Only work done along the odd segments contribute ๏ฟฝโ๏ฟฝ ๐2๐โ1 โ โ๐2๐โ1=
๐น ๐2๐โ1 โ โ๐2๐โ1
lim๐โโ
๏ฟฝ ๐น(๐)๐๐
๐๐
๐๐
โก ๏ฟฝ๐น ๐2๐โ1 โ โ๐2๐โ1
๐
๐=1
๐๐ =
In the limit ๐ต โ โ 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
๐น ๐ = โ๐(๐ โ ๐ฟ0)
Work done by a Spring ๐๐ โก ๏ฟฝ ๐น๐ ๐ ๐๐
๐๐
๐๐
= ๏ฟฝ โ๐ ๐ โ ๐ฟ0 ๐๐
๐๐
๐๐
= โ๐ ๏ฟฝ ๐ โ ๐ฟ0 ๐๐
๐๐
๐๐
Change of Variable ๐ฅ = ๐ โ ๐ฟ0, ๐ฅ = ๐๐
๐๐ โก โ๐ ๏ฟฝ ๐ฅ๐๐ฅ
๐๐โ๐ฟ0
๐๐โ๐ฟ0
= โ12๐๐ฅ
2๐๐โ๐ฟ0
๐๐โ๐ฟ0
โ ๐๐ = โ12๐ ๐๐ โ ๐ฟ02 โ ๐๐ โ ๐ฟ0 2
Or of we let ๐ represent the deformation (which we already did), then
๐๐ = โ12๐ โ๐ฟ๐
2 โ โ๐ฟ๐ 2 , or ๐๐ = โ12๐ ๐ฅ๐2 โ ๐ฅ๐2
๐๐ ๐ โ ๐๐ ๐ฟ0 = โ๐๐ ๐ฟ0โ๐ = โ โ 12๐ ๐ โ ๐ฟ0 2 โ ๐ฟ0 โ ๐ฟ0 2
= + 12๐ ๐ โ ๐ฟ0 2 โ 0 2 = 12๐ ๐ โ ๐ฟ0 2
Taking the relaxed spring to have ZERO potential energy (usually best choice)
๐๐ = 12๐ โ๐ฟ 2, or ๐๐ = 1
2๐๐ฅ2
Potential Energy of a Spring
NOTE: ๐ is not spring length
NOTE: ๐ 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: ๐ is not spring length
Example 08-02 (1/3) A block of mass ๐ is pushed up against a spring, compressing it a distance ๐ฅ, and the block is then released. The spring projects the block along a frictionless horizontal surface, giving the block a speed ๐ฃ. The same spring is then used to project a second block of mass 4๐, giving it a speed of 5๐ฃ. What distance ๐ฅ2 was the spring compressed in the second case? Answer in terms of a numerical factor times ๐ฅ, 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 ๐ , compression ๐ฅ, results in speed ๐ฃ. same spring on mass 4๐: speed of 5๐ฃ. What distance was the spring compressed in the second case? i.e. ๐ฅ2/ ๐ฅ = ?
(%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 ๐ , compression ๐ฅ, results in speed ๐ฃ. same spring on mass 4๐: speed of 5๐ฃ. What distance was the spring compressed in the second case? i.e. ๐ฅ2/ ๐ฅ = ?
(%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)
๐๐บ = ๏ฟฝ ๐น๐ ๐ ๐๐
๐๐
๐๐
= ๏ฟฝ โ๐บ๐บ๐๐2
๐๐
๐๐
๐๐
= โ๐บ๐บ๐ ๏ฟฝ ๐โ2๐๐
๐๐
๐๐
= โ๐บ๐บ๐โ1
๐โ1 ๐๐๐๐
๐๐บ = ๐บ๐บ๐1๐๐โ
1๐๐
FG
M
m ๐
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
๐น๐ = โ๐บ๐บ๐๐2
The minus sign means it points inward
It is conventional to choose ๐G = 0 at ๐ = โ
๐G = 0 โ ๐บ๐บ๐1๐โ
1โ
=โ๐บ๐บ๐
๐
๐G =โ๐บ๐บ๐
๐
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 instructorโ terms:
โ๐ธ = ๏ฟฝ work done by non โ conservative force ๐๐
Unit 09