PHYS 151 Fall 2016
Supplemental Instruction
http:/www.usc.edu/si
Alyssa Tsenter
[email protected]
Final Exam Review
A brief review of the new material:
Gravitation:
Newton proposed a force law that states that every particle attracts any other par
PHYS 151 Fall 2016
Supplemental Instruction
http:/www.usc.edu/si
Alyssa Tsenter
[email protected]
Midterm 2 Exam Review
A brief review of the material:
Centripetal Forces:
If there is a force causing UCM, then:
&' (
=
becomes
% =
= +
)
Note: Fc
PHYS 151 Fall 2016
Supplemental Instruction
http:/www.usc.edu/si
Alyssa Tsenter
[email protected]
Midterm 1 Exam Review
A brief review of the material:
Dimensional Analysis:
Use to approximately solve problems without going through all the work of
rn1 Sn = rn Sn which is Harmonic. 6.16 Surface density in terms of
surface Harmonics The potential at any point P due to a number of
particles situated on the surface of sphere of radius a can be ut in the
form V1 = = + n 0 n 1 n a r Un , when r < a (1) V
particle of mass m moves in a force field of potential V. Write the
Hamiltonian. Solution:- Here K.E. is T = ( ) 2 2 2 m x y z 2 1 & + & + &
m p , z m p , y m p x x y z &= &= &= and P.E. is V (x,y,z) H = T +
V H = (p p p ) V( ) x, y,z 2m 1 2 z 2 y 2 x +
, and the differential equation (26) becomes a simple harmonic oscillator
equation with solution .t / 0 cos.!t C /; (27) where 0 is the (angular)
amplitude of the pendulum, ! D pg=` is the angular frequency, and is a
phase constant that depends on where t
two coupled first-order ordinary differential equations, which may be
solved simultaneously to find .t / and L.t /. Note that for this example,
Eq. (45) is equivalent to L D I!, and Eq. (48) is the torque D mg` sin . 6
GAMEPLAY AND GAME MECHANICS DESIGN G
= 0 to x = 2a (here r = a) 108 Then V = 2a 0 dx a 2 a = a M a 4 a 2
= Case(iii) When P is inside the spherical shell, limit are from x = (a
r) to (a + r) V = 4a = a M 6.6 Attraction of a spherical shell Let us
consider a slice BBCC at point P , the attra
space remains Invariant Proof :- The position of a point on any 2 D
surface is specified completely by two parameters, e.g. u, v Then ( ) ( )
= = p p u,v q q u,v i i i i (2) In order to transform integral (1) into
new variables (u, v), we take the relati
part (Hanging from wall), since the force doesnt change we have the
same force as the part one. But, now our area is changed and as you can
see in Figure (5 b) the force is parallel to the area. It means now we have
shear stress, which we can find it with
156.2 = E = = 156.2 200 = 0.781 103 =
= = 0.781 103 0.7 = 0.546 103 = 0.546
Figure 9. Engine block hanging from cable [1] 13 | S o l i d M e c h a n i
c s Stress-Strain Diagram Earlier we mentioned about the stress-strain
diagram. This is a graph of st
3 1 a 2 (2) from (1) & (2), we have = 4 3 [MA2 MP2 ] = 4 3
(MA + MP) (MA MP) = 4 3 (PB) (AP) At P on OA (x = 3 y) =
n V 4 1 1 = x 3 y 1 1 y V cos 30 x V sin 30 4 1 =
+ = x 3 y 1 1 y V 2 3 x V 2 1 4 1 = +
= x 3y 2 2 y ax 3 3xy 3 3ay 2 3 x 2 3 4 1 = +
(MPa) * 36 (mm2) = 240 (106 2 ) * 36 (10-6 m2) = 8640 N = 8.64
kN 15 | S o l i d M e c h a n i c s Stress concentration Airplane window
(de Havilland Comet)
http:/en.wikipedia.org/wiki/De_Havilland_Comet#Comet_disasters_of_
1954 Up to now we assumed that
course how to take the derivative of a function of one variable. For
example, if f .x/ D 3x2 C 7x5 (3) then df dx D 6x C 35x4: (4) But what
if f is a function of more that one variable? For example, if f .x; y/ D
5x3y5 C 4y2 7xy6 (5) then how do we take t
instead of Newtonian mechanics is sometimes advantageous in certain
problems, where the equations of Newtonian mechanics would be quite
difficult to solve. In Lagrangian mechanics, we begin by defining a
quantity called the Lagrangian (L), which is define
3 External forces for tension and compression tension compression F 5 |
S o l i d M e c h a n i c s A positive sign for stress will be used to
indicate a tensile stress (member in tension) and a negative sign to
indicate a compressive stress (member in co
in axial direction, as shown in figure 12. In this figure w = 20 cm, d = 2
cm, h = 4 cm. Find the maximum stress in the bar (Kt = 2.5) Solution:
The cross section with the hole has the less area and because of
geometry change there is stress concentration
yi, zi are the components of i r i.e. ) Fi = (Xi ,Yi ,Zi and ) r
(x ,y ,z i = i i i Then W = = + + N i 1 i i i i i i ( . If the system is
Holonomic ,i.e., the co- X x Y y Z z ) ordinate qj changes to qj + qj
without making any change in other (n1) coordin
(8) Qj = = + N i 1 j i i i q r (F F ) Qj = =
+ N i 1 j i i i q r (F F ) (9) from (6) & (9), we get j j j Q q T q T
dt d = & , j = 1, 2,n This is a system of n
equations known as Lagrange equations. 45 S(fixed) Initial line 2 r m
P(r,) (m) P Planet S Sun
2 2 v P q . p F v q q . q F Now, ( ) ( ) = i i i i i i i i
v p v q u p u q u, v q , p = + +
i,k k i k 2 2 k i k 2 i k i k 2 2 k i k 2 2 i v P q P F v q q q F v q u P q P
F u q q q F u q = +
i k ,k i, k i k 2 2 i k i k 2 2 i k i k 2 2 i k i k 2 2 i v P
126 = n (n + 1) rn Sn (1) n (n + 1) rn Sn + 2 n 2 n 2 n n S r sin 1
S sin.r sin 1 + = 0 n(n + 1) rn Sn + 0 S
sin r S sin sin r 2 n 2 2 n n n = + n(n
+1) Sn + 0 S sin 1 S sin sin 1 2 n 2 2 n = +
n(n +1) Sn + cot 0 S sin 1 S S 2 n 2 2 2 n 2 n = + +
(2)
cos p cos ec d r 2mE 2 2 2 1 1 + p + constant 5.7 Lagranges
Brackets Lagranges bracket of (u, v) w.r.t. the basis (qj, pj) is defined as
cfw_u, vq,p or (u, v)q,p = j j j j j v q u p v p .
u q Properties: (1) (u, v) = (v, u) (2) (qi, qj) = 0 (3) (pi, pj) =
m r mr & & = (3) m 2 2 r m &r& mr & = 2 2 r r r
& & = and (4) (r ) 0 dt d 2 & = 3.5 Lagrange equation for a
conservative system of forces Suppose that the forces are conservative &
the system is specified by the generalized co-ordinate qj (j = 1, 2,.n).
recalling that it should be expressed in radians, we write: tan =
0.04 4 = 0.01 F 4 in. 8 in 2 in. 0.04 in. C B A Figure 8. Shear
strain 11 | S o l i d M e c h a n i c s Hookes Law In 1678 the English
mathematician Robert Hooke (16351703) stated a law whi
+ 2E K sin q K m t 1 ( ) + = q 2E K t sin
m K 1 1 ( ) = + 1 t m K q sin 2E k q(E, t) = ( ) +
1 t m K sin K 2E The constants 1, E can be found from initial
conditions The momentum is given by p = ( ) q S (t) V(q) q S 1 + =
= q V = 2 1 2 Kq 2 1 2m E = (
= dx PQ m V = d psec mpsec 2 = m secd = m
+ 4 2 log tan = m +
+ 4 2 log tan 4 2 log tan = m log
+ + 2 4 tan 2 4 tan 6.3 Potential at a point P on
the axis of a Uniform circular disc or plate:- We consider a uniform
circular disc of radius a & P is a
Using the Method of separation of variable, we have S = S1(t) + R(r) +
() V(r) d d r 1 dr dR 2m 1 dt dS 2 2 2 1
+ = L.H.S. is function of t and R.H.S. is function of r &
but not of t, therefore it is not possible only where each is equal to
constant =
(pj)t= (pj)t=0 = + 0 0 j j dt Q q T dt Let Qj, 0 in such
a way that 49 = 0 j j 0 Qj Lim (finite) (j = 1, 2,., n) Q dt J
Further as the co-ordinate qj do not change suddenly, = 0 j 0
dt 0 q T Lt Writing pj = 0 Lt [(pj)t= (pj)t=0], We thus obtain
Lagranges
rod. Deformation usually denoted by the Greek letter delta (). Prior to
applying force the bar has a length L1. The applied tensile force causes
the bar to increase its length to L2. The difference between two lengths
(prior and after applying load) is de
k j i j q r q t q r q & [ k q& are independent of qj] = i n k 1 k k j r q
q q t & + = = (r ) dt q dr q i j i j & =
j i j i q r q r dt d = & (4) Also
we know that j i j i q r q r = & & (5) Consider
+ = j i i j i i j i i q r dt d r q r r q r r dt d &
&