PHY 3003 SPRING 2015: SOLUTIONS, PROBLEM SET 10 (1) 10.6

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PHY 3003 SPRING 2015: SOLUTIONS, PROBLEM SET 10
(1) 10.6
The mass is
b
Z
Z
2
M=
r dr
dcos(θ)dφρ =
a
so
2πρ 3
b − a3
3
3M
2π (b3 − a3 )
The center of mass must by symmetry be along the line X = Y = 0 so
Z b
Z
3 b4 − a4
πρ 4
ρ
r2 dr dcos(θ)dφ rcosθ =
b − a4 =
ZCOM =
M a
4M
8 b3 − a3
ρ=
When a = 0 we get 3b/8 and when a → b we get 3b/4.
(2) 10.12
√
The coordinates
of the
√
√ corners of the triangle in the x−y plane are (−a, −a/ 3),
(a, −a/ 3) and (0, 2a/ 3). Assume the height of√the prism is b Thus the equation
for say the left side of the triangle is y = − √a3 + 3(x + a) so that
0
Z
M
= 2ρb
Z
√
− √a + 3(x+a)
3
dx
−a
0
Z
dx
= 2ρb
dy
− √a
3
√
√
3(x + a) = 3ρ ba2
−a
(This shows we got the limits of integration right!)
Now
Z
0
Izz = 2ρb
Z
√
− √a + 3(x+a)
3
dx
−a
dy x2 + y 2
− √a
3
1 √
3 0
√
2
= 2ρb
dx x
3(x + a) +
3(x + a)
3
−a
Z 0
√
2M a2
2ρ
= 2 3ρ b
dxx2 (x + a) + (x + a)3 = √ ba4 =
3
3
−a
Z
The two products of inertia for rotations about the z axis must be zero because
for rotations about z the symmetry is high enough that the angular momentum is
parallel to z. Put differently, the center of mass is obviously on the z axis.
1
2
(3) 10.18 (a)
Change in angular momentum about pivot is the impulsive torque about the pivot,
in other words
∆L = ∆T = ξb
This implies the angular velocity is ω = ∆L/I = ξb/I which means that the
velocity of the center of mass is Vcom = aω so the center of mass momentum
PCOM = M aξb/I
(b) Now, the total angular momentum is the sum of the angular momentum
obtained by treating the system as a point particle rotating about the center of mass
plus the angular momentum of rotation about the center of mass. Thus the angular
momentum of rotation about the center of mass is
Lrot = ξb − M a2 ξb
Because the system is fixed at the pivot point, it is not actually rotating about
the center of mass immediately after the impulse, thus the pivot must deliver the
impulse needed to cancel Lrot , i.e. must deliver the impulsive torque
Tpivot = −Lrot = ξb
M a2
−1
I
To get the force, divide by the distance to the COM, a.
(c) The sweet spot is thus
r
b0 =
I
M
(4) 10.22
(a) Let m be the mass of one of the vertices. Choose coordinates so that the
corner about which we wish to find the inertia tensor is the origin and the other 7
points are at
~ 1 = aˆ
R
x
~ 2 = a (ˆ
R
x + yˆ)
~ 3 = aˆ
R
y
~ 4 = aˆ
R
z
~
R5 = a (ˆ
x + zˆ)
~ 6 = a (ˆ
R
x + yˆ + zˆ)
~ 7 = a (ˆ
R
y + zˆ)
3
so that
Ixx = 8ma2
Iyy = 8ma2
Izz = 8ma2
Ixy = −2ma2
Ixz = −2ma2
Iyz = −2ma2
(b) About cube center all of the off diagonal elements are zero by symmetry and
2
also by symmetry the diagonal elements are equal, with I = 8m a2 = 4ma2 .
(5) 10.32
The diagonal components of the inertia tensor (in any basis) are
Z 0
Izz =
d3 r ρ(r) x2 + y 2
while
Z
Ixx + Iyy =
0
d3 r ρ(r) y 2 + z 2 + x2 + z 2 = int0 d3 r ρ(r) x2 + y 2 + 2z 2 ≥ Izz
Note that this inequality holds in any basis, and thus holds in particular in the basis
in which the inertia tensor is diagonal.
Equality occurs only for a “pancake” (no extension in the z direction).
(6) 10.36
Note that the problem does not specify the origin. We will compute the moment
of inertia about the origin.
Ixx = ma2 (0 + 5 + 5) = 10ma2
Iyy = ma2 (1 + 4 + 1) = 6ma2
Izz = ma2 (1 + 4 + 1) = 6ma2
Ixy = −ma2 (0 + 0 + 0)
Ixz = −ma2 (0 + 0 + 0)
Iyz = ma2 (0 + 2 + 2) = −4ma2
or


10 0
0
6 −4 
I = ma2  0
0 −4 6
4
The principal axes are
e1 = x
ˆ
1
y + zˆ)
e2 = √ (ˆ
2
1
e3 = √ (ˆ
y − zˆ)
2
and in this basis the matrix is


10 0 0
I = ma2  0 2 0 
0 0 10

PHY 3003 SPRING 2015: SOLUTIONS, PROBLEM SET 10 (1) 10.6

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