Solutions for Homework #8
11-2.
a)
Moments of inertia with respect to the xi axes:
x3 = x3′
R h
CM
x2′
x1′
x2
x1
It is easily seen that Iij = 0 for i ≠ j. Then the diagonal elements Iii become the principal
moments Ii, which we now calculate.
The computation can be simplified by noting that because of the symmetry, I1 = I2 ≠ I3 .
Then,
I1 = I2 =
I1 + I2 ρ
= ∫ 2x32 + x12 + x22 dv
2
2
(
)
(1)
which, in cylindrical coordinates, can be written as
I1 = I2 =
ρ
2∫
2π
0
dφ ∫ dz ∫
h
Rz h
0
0
( r + 2z ) rdr
2
2
(2)
where
ρ=
M
3M
=
V π R 2h
(3)
Performing the integration and substituting for ρ, we find
I1 = I2 =
(
3
M R 2 + 4h2
20
)
(4)
I3 is given by
(
)
I3 = ρ ∫ x12 + x22 dv = ρ ∫ r2 ⋅ rdrdφ dz
(5)
from which
I3 =
b)
3
M R2
10
Moments of inertia with respect to the x′i axes:
Because of the symmetry of the body, the center of mass lies on the x′3 axis. The
coordinates of the center of mass are (0,0,z0 ), where
(6)
z0 =
∫ x′ dv = 3 h
∫ dv 4
3
(7)
Then, using Eq. (11.49),
⎤
Ii′j = Iij − M ⎡⎣ a2δ ij − aa
i j⎦
(8)
In the present case, a1 = a2 = 0 and a3 = ( 3 4) h , so that
I1′ = I1 −
9
3
1 ⎞
⎛
M h2 =
M ⎜ R 2 + h2 ⎟
16
20 ⎝
4 ⎠
I2′ = I2 −
9
3
1 ⎞
⎛
M h2 =
M ⎜ R 2 + h2 ⎟
⎝
16
20
4 ⎠
I3′ = I3 −
3
M R2
10
11-7.
x
θ
R
r
d
φ
The force between the force center and the disk is, from the figure
F = − kr
(1)
Only the component along x does any work, so that the effective force is
Fx = − kr sin φ = − kx . This corresponds to a potential U = kx2 2 . The kinetic energy of the
disk is
T=
1
1
3
M x& 2 + Iθ& 2 = M x& 2
2
2
4
(2)
where we use the result I= M R 2 2 for a disk and dx = R dθ. Lagrange’s equations give
us
3
&& + kx = 0
Mx
2
This is simple harmonic motion about x = 0 with an angular frequency of oscillations
ω=
2k
3m
(3)
11-11.
a)
No sliding:
2
2
P
P
From energy conservation, we have
mg
l
l 1
1
= m g + m vC2 .M . + Iω 2
2 2
2
2
(1)
where vC M is the velocity of the center of mass when one face strikes the plane; vC .M . is
related to ω by
vC M =
l
ω
2
(2)
I is the moment of inertia of the cube with respect to the axis which is perpendicular to
one face and passes the center:
I=
1 2
ml
6
(3)
Then, (1) becomes
m gl
2
(
)
2−1 =
1
m
2
2
⎡ lω ⎤ 1 ⎡ m l ⎤ 2 1 2 2
+
⎢⎣ 2 ⎥⎦ 2 ⎢ 6 ⎥ ω = 3 m l ω
⎣
⎦
2
(4)
from which, we have
ω2 =
b)
3g
2l
(
)
2−1
(5)
Sliding without friction:
In this case there is no external force along the horizontal direction; therefore, the cube
slides so that the center of mass falls directly downward along a vertical line.
θ
P
P
While the cube is falling, the distance between the center of mass and the plane is given
by
y=
l
cosθ
2
Therefore, the velocity of center of mass when one face strikes the plane is
(6)
=−
y&
l
sin θ θ&
2
0= π 4
1
1
= − lθ& = − lω
2
2
(7)
θ =π 4
From conservation of energy, we have
2
mg
l
l 1 ⎛ 1 ⎞
1⎛ 1
⎞
= m g + m ⎜ − lω ⎟ + ⎜ m l 2 ⎟ ω 2
⎠
2 2 ⎝ 2 ⎠
2⎝ 6
2
(8)
from which we have
ω2 =
12 g
5 l
(
)
2−1
(9)
Bonus
11-5.
z
M
J
z–a
a
Q
a) The solid ball receives an impulse J; that is, a force F(t) is applied during a short
interval of time τ so that
J= ∫ F ( t′ ) dt′
(1)
dp
=F
dt
(2)
dL
= r× F
dt
(3)
Δp = ∫ F ( t′ ) dt′ = J
(4)
ΔL = ∫ r× F ( t′ ) dt′ = r× J
(5)
The equations of motion are
which, for this case, yield
Since p(t = 0) = 0 and L(t = 0) = 0, after the application of the impulse, we have
p = M V C M = J; L = I0 ω = r× J= ( z − a) J
so that
ω
ω
(6)
V CM =
J
M
(7)
and
ω=
J
ω
( z − a)
ω
I0
(8)
where I0 = ( 2 5) M a2 .
The velocity of any point a on the ball is given by Eq. (11.1):
vα = V C M + ω × rα
(9)
For the point of contact Q, this becomes
vQ = V C M − ω a
=
J
J
J ⎡ 5( z − a) ⎤
1−
M ⎢⎣
2a ⎥⎦
(10)
Then, for rolling without slipping, vQ = 0 , and we have
2a = 5( z − a)
(11)
so that
z=
7
a
5
(12)