PHYSICS 110A : CLASSICAL MECHANICS FINAL EXAM SOLUTIONS [1] Two blocks and three springs are configured as in Fig. 1. All motion is horizontal. When the blocks are at rest, all springs are unstretched.
Figure 1: A system of masses and springs. (a) Choose
110a Homework 5 solutions
1. Taylor 8.2.
2
2
1
L = 2 m1 r1 + 1 m2 r2 (m1 gz1 + m2 gz2 + U (r ) = Lcm + Lrel
2
2
1
2
with Lcm = 2 M R M gZ and Lrel = 1 r U (r ). The CM coordinate R moves with zero
2
acceleration in the X and Y components, and with
d2 Z
dt
PHYSICS 110A : CLASSICAL MECHANICS MIDTERM EXAM #2 [1] Two blocks connected by a spring of spring constant k are free to slide frictionlessly along a horizontal surface, as shown in Fig. 1. The unstretched length of the spring is a.
Figure 1: Two masses c
Chapter 3
One-Dimensional Conservative Systems
3.1 Description as a Dynamical System
For one-dimensional mechanical systems, Newtons second law reads m = F (x) . x (3.1)
A system is conservative if the force is derivable from a potential: F = dU/dx. The t
Chapter 2
Systems of Particles
2.1 Work-Energy Theorem
Consider a system of many particles, with positions ri and velocities ri . The kinetic energy of this system is 1 2 T = (2.1) Ti = 2 mi ri .
i i
Now let's consider how the kinetic energy of the syste
Physics 110A Midterm 2
Dec. 5, 2014
PROBLEM (1)
(10 points)
Find the curve y(x), that goes between points (x = 0, y = 0) and (x = 1, y = 1), and that minimizes
1
J = [y 2 2y 2 ]dx.
0
f = y 2 2y 2
d
4y 2y = 0
dx
y = 2y
y = A sin( 2x) + B cos( 2x)
y = A sin
Physics 110A Final Exam
Dec. 17, 2014
PROBLEM (1)
(6 points)
A sphere of uniform density, radius R and mass M is assembled by bringing mass in from innity.
Calculate the work done by gravity during the assembly of such a sphere. Verify that your answer
ha
Vectors and Coordinates
ei ej = ij
3
AB =
Ai Bi = AB cos(AB )
i=1
ds2 = dr2 + r2 d2 + r2 sin2 d2
3
(A B)i =
ijk Aj Bk
ijk lmk
j,k
x = r sin cos
ds2 = dr2 + r2 d2 + dz 2
= il jm im jl
k=1
y = r sin sin
x = r cos
z = r cos
y = r sin
Newtons Laws and
Physics 110A Midterm 2
Dec. 5, 2014
PROBLEM (1)
(10 points)
PROBLEM (2)
(10 points)
PROBLEM (3)
(10 points)
Vectors, Coordinates, Identities, Integrals.
3
(A B)i =
AB =
ei ej = ij
ijk Aj Bk
Ai Bi = AB cos(AB )
i=1
j,k
3
ds2 = dr2 + r2 d2 + r2 sin2 d2
ij
Chapter 4
Linear Oscillations
Harmonic motion is ubiquitous in Physics. The reason is that any potential energy function, when expanded in a Taylor series in the vicinity of a local minimum, is a harmonic function:
U (q )=0 N N q=q (qj - qj ) + 1 2 j,k=1
Chapter 5
Calculus of Variations
5.1 Snell's Law
Warm-up problem: You are standing at point (x1 , y1 ) on the beach and you want to get to a point (x2 , y2 ) in the water, a few meters offshore. The interface between the beach and the water lies at x = 0.
Chapter 7
Noethers Theorem
7.1 Continuous Symmetry Implies Conserved Charges
Consider a particle moving in two dimensions under the inuence of an external potential U (r). The potential is a function only of the magnitude of the vector r. The Lagrangian i
INSTRUCTIONS:
Physics 110A: Final Exam
Do four problems.
of problems 3, 4, and 5.
You must do problems 1 and. 2. Then choose any two
Central Forces
Two particles attract each other via a central force 170-) = Cr3/2, where C. is a constant.
(8) Write
Physics 110A: Midterm Exam #1
[1] A damped harmonic oscillator obeys the dimensionless equation of motion, x + 8 x + 36 x = 0 . (a) Write down the solution x(t) under the conditions x(0) = 1 and x(0) = -3. (b) Let the oscillator now be forced at (dimensio
Physics 110A: Final Exam
INSTRUCTIONS:
Do problems 1, 2, and 3 and either 4 or 5.
[1] A particle of mass m moves in these one-dimensional potentials: (a) U (x) = U0 |x/a| ( > 0), (b) U (x) = -U0 sech 2 (x/a), and (c) U (x) = U0 tan2 (x/a). In all cases, U
Physics 110A: Final Exam INSTRUCTIONS: Do four problems. of problems 3, 4, and 5. [1] Central Forces Two particles attract each other via a central force U (r) = -Cr-3/2 , where C is a constant. (a) Write down and sketch the effective potential Ueff (r).
110a Homework 6 solutions
1. Taylor 8.11.
2. Taylor 8.31.
For U = k/r with k > 0, get that Uef f > 0, so E > 0. To go from the case in the book
to this case, just take k k , which takes c c and does not change the expression
relating E and
(so
> 1 since E
110a Homework 4, solutions
1. Taylor 7.14.
1
1
3
3
Use I = 2 mR2 and = x/R. Get T = 1 mx2 + 2 I 2 = 4 mx2 , so L = 4 x2 + mgx
2
and the EL equation yields
d2 x
dt2
= 2g/3.
With lagrange multipliers now. L = 1 mx2 + 1 I 2 + mgx + (x R ). Get
2
2
m
d2 x
= m
110a Homework 1 solution sketches
1. straightforward (hopefully you worked out the details!).
2. Inertial frame: = 0, r = R v0 t. Rotating frame: r = r , = t.
3. (a)
v
v0
mev/V dv = F0
t
dt =
0
v0 /V
(b) At rest at tr = (1 e
F0 t, so v = V ln(F0 t/mV + ev