Full file at
https://buklibry.com/download/solutions-manual-classical-mechanics-with-problems-and-solutions-1st-edition-by-david-morin/
SOLUTIONS MANUAL
Introduction to Classical Mechanics
With Problems and Solutions
David Morin
Cambridge University Press
Download full file from buklibry.com
Full file at
https://buklibry.com/download/solutions-manual-classical-mechanics-with-problems-and-solutions-1st-edition-by-david-morin/
TO THE INSTRUCTOR: I have tried to pay as much attention to detail in
these exercise solutions as I did in the problem solutions in the text. But
despite working through each solution numerous times during the various
stages of completion, there are bound to be errors. So please let me know if
anything looks amiss.
Also, to keep this pdf file from escaping to the web, PLEASE don’t distribute
it to anyone, with the exception of your teaching assistants. And please make
sure they also agree to this. Once this file gets free, there’s no going back.
In addition to any comments you have on these solutions, I welcome any
comments on the book in general. I hope you’re enjoying using it!
David Morin
morin@physics.harvard.edu
(Version 2, April 2008)
c David Morin 2008
°
Download full file from buklibry.com
Full file at
https://buklibry.com/download/solutions-manual-classical-mechanics-with-problems-and-solutions-1st-edition-by-david-morin/
48
CHAPTER 5. CONSERVATION OF ENERGY AND MOMENTUM
(b) This only difference is that now the integral starts at L/2 instead of zero. So
we have
F = 2πσGm sin θ cos θ
Z
L
L/2
dx
= 2πσGm sin θ cos θ (ln 2).
x
(176)
Since sin θ cos θ = (1/2) sin 2θ, this force is maximized when 2θ = 90◦ =⇒ θ =
45◦ .
5.58. Sphere and cones
(a) There is no change in speed inside the shell. The potential energy at the surface
of the shell is
V (R) = −
Gm(4πR2 σ)
GmM
=−
= −4πGmRσ.
R
R
Conservation of energy then gives 0 = mv 2 /2 − 4πGmRσ =⇒ v =
(177)
√
8πGRσ.
(b) Let’s find the potential energy at the tip of the cones, due to one of the cones.
We’ll slice the cone into rings and then integrate. Consider a thin ring around
the cone, located at a slant distance x away from the tip. The radius r of the
ring is given by r/x = R/L =⇒ r = xR/L. So
¡
¢
Gm 2π(xR/L) dx σ
Gm dM
=−
= −2πGm(R/L)σ dx.
dV = −
x
x
(178)
Integrating from x = 0 to x = L simply gives V = −2πGmRσ. We need to
double this because there are two cones, so we end up with the same potential
of −4πGmRσ
as in part (a), which means that we obtain the same speed of
√
v = 8πGRσ, independent of L.
5.59. Ratio of potentials
In the first picture, the big square can be built up from four of the small ones (with
the mass at the corner of each), so A = 4.
In the second picture, consider a tiny patch of area in the small square. This
patch gives some contribution to the potential energy of m. Now consider the
corresponding patch in the big square. What is the contribution of this patch to the
potential energy of m? Well, the larger patch has four times the area (and hence
mass) as the smaller patch, because areas are proportional to lengths squared. But it
is also twice as far from m, compared with how far the smaller patch is from m in the
small square. So if the smaller patch contributes Gm(dM )/r to the potential in the
smaller square, then the larger patch contributes Gm(4dM )/2r to the potential in
the larger square. This is twice as much, and this relation holds for all corresponding
patches, so B = 2.
Putting the two pictures together then tells us that a mass at the center of a given
square has twice the (negative) potential that a mass at a corner of the same square
has.
5.60. Solar escape velocity
Let v0 ≈ 30 km/s be the orbital speed of the earth. Let the desired speed
p with respect
to the earth be v. The escape velocity from just the earth is ve = 2GMe /Re ≈
11.2 km/s. The escape velocity from the sun, starting at the location
p of the earth’s
2GMs /Res ≈
orbit (but excluding the orbital motion of the earth), is vs =
42 km/s, where Res is the earth-sun distance.
After the object has escaped the earth’s gravitational
field, conservation
of energy
p
√
v 2 − 2GMe /Re = v 2 − ve2 . The
gives the speed with respect to the earth as
speed with respect to the sun at this point is then (assuming√that the object is
wisely fired along the direction of the earth’s orbital motion) v 2 − ve2 + v0 . By
Download full file from buklibry.com
Full file at
https://buklibry.com/download/solutions-manual-classical-mechanics-with-problems-and-solutions-1st-edition-by-david-morin/
98
CHAPTER 8. ANGULAR MOMENTUM, PART I (CONSTANT L̂)
8.59. No final rotation
Let vb and vs be the final velocities of the ball and stick, with rightward and leftward
taken to be positive, respectively. Conservation of p, E, and L (around a dot on the
table where the initial center of the stick is) give
0
2
1 mℓ
ω02
2 12
³ 2´
mℓ
ω0
12
³
´
=
M vb − mvs ,
´
1
1
0 + mvs2 + M vb2 ,
2
2
ℓ
(0 + 0) + M vb .
2
³
=
=
(455)
(We have noted that in general the stick’s contributions come from both the motion
of the CM and the motion relative to the CM.) Using the first and third equations
to write vb and vs in terms of ω0 , the second equation gives M = m/2.
8.60. Same final speeds
Let v be the common final speed. Let ω be the final angular speed of the stick.
Then conservation of p, E, and L (around a dot on the table where the center of
the stick is when the collision occurs) give
mv0
1
mv02
2
0
=
=
=
2mv,
³
´
1
1
1
(Amℓ2 )ω 2 + mv 2 + mv 2 ,
2
2
2
ℓ
−(Amℓ2 )ω + mv .
2
(456)
The first equation gives v = v0 /2, and then the third equation gives ω = v/2Aℓ =
v0 /4Aℓ. Plugging these into the second equation yields A = 1/8.
8.61. Perpendicular deflection
Let vx and vy be the final velocity components of the center of the dumbbell, with
rightward and downward positive, respectively. The moment of inertia of the dumbbell around its center is 2m(ℓ/2)2 = mℓ2 /2. Conservation of px , py , E, and L
(around a dot on the table where the initial center of the stick is) give
M V0
=
0
=
1
M V02
2
=
ℓ
2
=
M V0
M V0
,
2m
Mu
,
M u − (2m)vy =⇒ vy =
2m
µ
³
´ ¶
1
1 mℓ2
1
ω2 ,
M u2 +
(2m)(vx2 + vy2 ) +
2
2
2
2
0 + (2m)vx
³
mℓ2
ω
2
´
=⇒ vx =
=⇒ ω =
M V0
.
mℓ
(457)
Plugging the vx , vy , and ω from the first, second, and fourth equations into the third
gives
u = V0
r
2m − 2M
.
2m + M
(458)
So we need m ≥ M for this setup to be possible (although the case of equality leads
to M being at rest).
8.62. Glancing off a stick
Since there is no force in the y direction
√on the mass (because the stick is frictionless),
the y speed of the mass remains v0 / 2, and the CM of the stick ends up moving
only in the x direction. Let v be the resulting speed of the CM, and let ω be the
Download full file from buklibry.com