7A Midterm II Sample Questions  Solutions
Kevin Young
This is a list of sample problems that I will be solving for the midterm review on
Saturday, 122 in 1 Pimentel. I may not get to solve all of these in the time allotted,
and I will most likely be doing them out of order.
Problem 1  Semicircular rod
A rod of mass, m, length,
π
r, is bent into a semicircle.
1. Calculate the center of mass of the rod after it has been bent. Use symme
try considerations to reduce the problem to onedimension. I would suggest
polar coordinates for this integral.
This problem is, like it suggests, best solved in polar coordinates.
Be
cause of the symmetry of the problem, the axis must lie along the ˆ
y
axis. Now,
this reduces the problem to finding the center of mass in the ˆ
y
direction. The
formula for the center of mass is:
¯
y
=
ydm
M
tot
Where here
dm
=
λ
ds
=
λ
r d
θ
, and
λ
=
M
tot
/(
leng th
)
=
M
tot
/(
π
r
) is the lin
ear mass density. And in polar coordinates,
y
=
r
sin
θ
. The integral becomes:
¯
y
=
y dm
M
tot
=
λ
r
2
sin
θ
d
θ
M
tot
=
λ
r
2
M
tot
π
0
sin
θ
d
θ
=
M
tot
r
2
M
tot
π
r
(

cos
θ
)
π
0
=
2
r
π
2. Calculate the moment of inertia about an axis perpendicular to the plane of
the semicircle and passing through the point midway between the ends of the
rod. (This is like a ring spun about its center of mass, but with the bottom half
cut away)
1
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No calculation needs to be done here, as all of the mass is located the
same distance (
r
) from the axis of rotation. So the moment of inertia is just
I
=
mr
2
3. Calculate the moment of inertia about and axis perpendicular to the plane of
the semicircle and through the center of mass. Don’t do any integrals for this
part.
This is occasion for the parallel axis theorem. We know where the center
of mass is, and we know what the moment of inertia through a different point
is, so the parallel axis theorem tells us that:
I
=
I
cm
+
md
2
Or, cast into a more useful form:
I
cm
=
I

md
2
Where here, d is the distance from the center of mass to the center of curvature
of the ring,
d
=
2
r
π
. Then the moment of inertia is:
I
=
mr
2

m
(
2
r
π
)
2
=
(1

4/
π
2
)
mr
2
.
Problem 2  Midspace Collision
Two identical satellites are in orbit around the Earth. The Earth has mass 6
×
10
24
kg
and radius 6400
km
. The gravitational constant is
G
=
6.67
×
10

11
Nm
2
kg

2
.
1. One of the satellites is in geosynchronous orbit. Find the distance the satellite
is from the center of the Earth and its linear speed.
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 Spring '08
 Lanzara
 Force, Mass, Potential Energy

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