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PHY 3003 SPRING 2015: WEEK OF MARCH 30
Reading: Taylor, Chapter 7.10, Chapter 9.1, 9.3, 9.4
NOTE EXAM in class Wednesday April 1. Covers material up through
Taylor Chapter 8 except for Lagrange multipliers
Homework: Due in class April 6. All problems count equally towards your grade. (Numbers refer to problems in Classical Mechanics, J. R. Talyor, 2005 Edition).
(1) 7.50 The Lagrangian is (note y increases downwards so the potential energy is
−mgy):
L=
m1 2 m2 2
x˙ +
y˙ + m2 gy + λ (x + y − L)
2
2
so we have
m1 x
¨ = λ
m2 y¨ = m2 g + λ
x+y = L
From the third equation we have x
¨ = −¨
y so λ = m1 x
¨ = −m1 y¨ and
(m2 + m1 ) y¨ = m2 g
or
y(t) = y0 + vy t +
and
λ=−
m2 g
t2
2 (m1 + m2 )
m1 m2
g
m1 + m2
(2) 7.52
L=
so
m 2 I ˙2
x˙ + φ + mgx + λ (x + Rφ)
2
2
m¨
x = mg + λ
I φ¨ = λR
x = −Rφ
The last equation implies
so
x
¨
φ¨ = −
R
λ=−
1
Ix
¨
R2
2
PHY 3003 SPRING 2015: WEEK OF MARCH 30
so
x
¨=
mg
m + RI2
and
mg
I
2
R m + RI2
Solving Newton’s equations in the standard way shows that λR is the torque exerted
by the tension in the rope on the wheel.
(3) 9.2
λ=−
We need ω such that
ω2R = g
i.e.
r
g
10
ω=
=
= 0.5/sec
R
40
Changing R from 40 to 38 is about 2 parts out of 40 or 5% change, so δgef f = ω 2 δR
so δg/g = δR/R = 0.05.
(4) 9.12
(a) If nothing is moving (in the rotating frame) then one simply has the extra
force ω 2 R (eq. 9.36)
(b) From the point of view of an observer in the rotating frame there is a radial
force F = ω 2 R which must be balanced by the force of static friction µmg so
µmg
Rmax = 2
ω
r
(5) 9.14
The easy way to approach this is to note that in the rotating frame the centripetal
force is equivalent to a potential energy
1
Vcentripetal = − Ω2 R2
2
That the surface of the water is an equipotential implies that
1
V = mgz(R) − Ω2 R2 = const
2
or
Ω2 2
z=C+
R
2mg
which is the equation for a parabola.