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 οΏ½βοΏ½πΆ
οΏ½ οΏ½βοΏ½πΆ β ππ β‘ 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 πΉπ΄) 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 ππ ) on that same object
The applied force ππ¨ is always in the direction of motion: πΎπ¨ > π over closed loop
The friction force ππ is always opposite the direction of motion: πΎππ < π over closed loop
Unit 07
NOTE: π is not spring length
This last point is confusingβ¦ donβt use it and donβt think about it. I prefer you remember the definition this way:
π π = π π0 βππ0βπ = βππ0βπ Example: gravity (CHOOSING π = 0 at π¦ = 0)
π π¦ = π 0 βπ0βπ¦ = 0 β βππ π¦ β 0 = πππ¦
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 +π¦ direction)
ππ = οΏ½βοΏ½π β βπ ππ = βππ βπ¦
π
ππ
ππ
π ππ
ππ ππ
ππ
π
ππ
ππ
π
ππ
ππ
π
ππ
ππ
Vertical component of force of gravity οΏ½βοΏ½π
Vertical component of displacement βπ
βπ¦<0 here
Gravity is a conservative force
Potential Energy of gravity ππ π¦ β ππ 0 = βππ = β βππ βπ¦
ππ π¦ = πππ¦ CHOOSE ππ = 0 at π¦ = 0 and UP to be +π¦
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 β’ All central forces are conservative β’ A central force is:
β one that is directed always towards or away from a βcenterβ (we usually assign this to be the origin)
β whose magnitude does not depend on the orientation β Whose magnitude depends only on the distance to the
βcenterβ
β’ Examples β Force exerted by a spring on a body tied to it at one end
(often the other end is tied to the Earth) β Force of gravity on a small object by another (often much
larger) object. β Electrical (actually: electrostatic) forces between charged
objects (PHYS 2220)
Frictionless, horizontal surface
π
Spring of force constant π
relaxed length πΏ0 π
ππ
ππ
Path from ππ to ππ
Approximate path by 2N that alternate π = 1,2,3, β¦π, (2π β 1)th: radial (2π)th: arc
οΏ½βοΏ½π β₯ βπ for arcs: work done along the even segments vanish
οΏ½βοΏ½ π2π β βπ2π= 0
οΏ½βοΏ½π β₯ βπ for radial steps: Only work done along the odd segments contribute οΏ½βοΏ½ π2πβ1 β βπ2πβ1=
πΉ π2πβ1 β βπ2πβ1
limπββ
οΏ½ πΉ(π)ππ
ππ
ππ
β‘ οΏ½πΉ π2πβ1 β βπ2πβ1
π
π=1
ππ =
In the limit π΅ β β 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
πΉ π = βπ(π β πΏ0)
Work done by a Spring ππ β‘ οΏ½ πΉπ π ππ
ππ
ππ
= οΏ½ βπ π β πΏ0 ππ
ππ
ππ
= βπ οΏ½ π β πΏ0 ππ
ππ
ππ
Change of Variable π₯ = π β πΏ0, π₯ = ππ
ππ β‘ βπ οΏ½ π₯ππ₯
ππβπΏ0
ππβπΏ0
= β12ππ₯
2ππβπΏ0
ππβπΏ0
β ππ = β12π ππ β πΏ02 β ππ β πΏ0 2
Or of we let π represent the deformation (which we already did), then
ππ = β12π βπΏπ
2 β βπΏπ 2 , or ππ = β12π π₯π2 β π₯π2
ππ π β ππ πΏ0 = βππ πΏ0βπ = β β 12π π β πΏ0 2 β πΏ0 β πΏ0 2
= + 12π π β πΏ0 2 β 0 2 = 12π π β πΏ0 2
Taking the relaxed spring to have ZERO potential energy (usually best choice)
ππ = 12π βπΏ 2, or ππ = 1
2ππ₯2
Potential Energy of a Spring
NOTE: π is not spring length
NOTE: π 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: π is not spring length
Example 08-02 (1/3) A block of mass π is pushed up against a spring, compressing it a distance π₯, and the block is then released. The spring projects the block along a frictionless horizontal surface, giving the block a speed π£. The same spring is then used to project a second block of mass 4π, giving it a speed of 5π£. What distance π₯2 was the spring compressed in the second case? Answer in terms of a numerical factor times π₯, 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 π , compression π₯, results in speed π£. same spring on mass 4π: speed of 5π£. What distance was the spring compressed in the second case? i.e. π₯2/ π₯ = ?
(%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 π , compression π₯, results in speed π£. same spring on mass 4π: speed of 5π£. What distance was the spring compressed in the second case? i.e. π₯2/ π₯ = ?
(%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)
ππΊ = οΏ½ πΉπ π ππ
ππ
ππ
= οΏ½ βπΊπΊππ2
ππ
ππ
ππ
= βπΊπΊπ οΏ½ πβ2ππ
ππ
ππ
= βπΊπΊπβ1
πβ1 ππππ
ππΊ = πΊπΊπ1ππβ
1ππ
FG
M
m π
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
πΉπ = βπΊπΊππ2
The minus sign means it points inward
It is conventional to choose πG = 0 at π = β
πG = 0 β πΊπΊπ1πβ
1β
=βπΊπΊπ
π
πG =βπΊπΊπ
π
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 instructorβ terms:
βπΈ = οΏ½ work done by non β conservative force ππ
Unit 09