Physics 105 Problem Set 3 Solutions
Problem 1
(3 pts)
a) We are asked to find the center of mass of a solid cone of mass
M
, height
L
and radius
R
. Namely, we
have to compute the following integral:
→
x
cm
=
cone
→
x dm
cone
dm
=
1
M
cone
→
x ρ
(
→
x
)
dV
In our case the cone has uniform density, so
ρ
=
M/V
. Let’s place the base of the cone on the
x

y
plane and the axis of the cone on the positive
z

axis. From rotational symmetry about the cone’s axis,
we know that the center of mass must be somewhere on the
z

axis, so the 3D problem can be reduced
to a 1D problem. The radius
r
of the cone as a function of
z
is
r
(
z
) =
R

zR/L
=
R
(1

z/L
). We need
to compute the following integral:
z
cm
=
1
M
cone
zdm
(
z
)
=
1
M
L
0
zρdV
(
z
)
=
1
M
M
V
L
0
zπr
(
z
)
2
dz
=
πR
2
V
L
0
z
(1

z/L
)
2
dz
=
πR
2
V
z
2
2

2
z
3
3
L
+
z
4
4
L
2
L
z
=0
=
πR
2
L
2
12
V
.
If you don’t happen to know that
V
(
cone
) =
1
3
base*height, you can calculate it by doing a similar integral
as above, just without the extra
z
in the integrand.
V
=
cone
dV
(
z
)
=
L
0
πr
(
z
)
2
dz
=
πR
L
0
(1

z/L
)
2
dz
=
πR
2
V
z

z
2
L
+
z
3
3
L
2
L
z
=0
=
πR
2
L
3
.
1
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Thus we have
z
cm
=
L/
4, which is independent of the radius
R
, and the 3D answer is
→
x
cm
= (0
,
0
, L/
4).
b) Now let us assume the cone is hollow, with a surface density of
σ
. Let’s split this problem up into two
parts, the upper part which looks like an upsidedown icecream cone and the base which is disc. We can
calculate the center of mass of each and then take the weighted average.
The surface area of the upper part can be found by integrating.
dl
is an infinitesimal segment tangent
to the surface of the cone, and
dz
is its vertical projection.
A
upper
=
upper
dA
(
z
)
=
upper
2
πr
(
z
)
dl
=
2
πR
L
0
(1

z/L
)
√
L
2
+
R
2
L
dz
=
2
πR
√
L
2
+
R
2
L
z

z
2
2
L
L
z
=0
=
πR
√
L
2
+
R
2
.
Now we can find its center of mass, which is similar to the integral above but with an extra
z
in the
integrand.
z
cm,upper
=
σ
M
upper
upper
zdA
(
z
)
=
1
A
upper
L
0
z
2
πR
(1

z/L
)
√
L
2
+
R
2
L
dz
=
1
πR
√
L
2
+
R
2
2
πR
√
L
2
+
R
2
L
z
2
2

z
3
3
L
L
z
=0
=
L
3
.
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 mechanics, Acceleration, Center Of Mass, Mass, Momentum, Velocity, reference frame

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