This preview shows pages 1–3. Sign up to view the full content.
Ch9HW2 – florin – (56950)
1
This printout should have 15 questions.
Multiplechoice questions may continue
on the next column or page – find all
choices before answering.
001 (part 1 of 2) 10.0 points
A man whose mass is 70 kg and a woman
whose mass is 55 kg sit at opposite ends
of a canoe 5 m long, whose mass is 45 kg.
Assume the man is seated at x = 0 and
the boat extends along the positive x axis
with the woman at the other end. Where is
the
center
of
mass
of
the
system
consisting of man, woman, and canoe?
Correct answer:
2.27941 m.
Explanation:
With the man (m
1
) at the origin, we have
x
CM
=
m
1
x
1
+
m
2
x
2
+
m
3
x
3
m
1
+ m
2
+ m
3
=
0 + (55 kg)(5 m) + (45 kg)(2.5 m)
70 kg + 55 kg + 45 kg
=
2.27941 m .
2
(part 2 of 2) 10.0 points
Suppose that the man moves quickly to
the center of the canoe and sits down
there. How far does the canoe move in the
water? Assume force of friction between
water and the canoe is negligible.
Correct answer:
1.02941 m.
Explanation:
Since friction is negligible, the net
horizontal
force on the system (Man+Woman+Canoe)
is
zero.
Therefore,
its
center
of
mass
momentum (and velocity) remain constant
(which is zero in this case). So the center of
mass of the system remains stationary.
For this problem, let us take the origin
to be at the left end of the canoe (note
that it is not a stationary point). The new
center of mass is at
x
CM
=
m
1
x
1
+
m
2
x
2
+
m
3
x
3
m
1
+ m
2
+ m
3
=
(70 kg)(2.5 m) + (55 kg)(5 m)
70 kg + 55 kg + 45
kg (45 kg)(2.5 m)
+
70 kg + 55 kg + 45 kg
=
3.30882 m .
At
first,
the
center
of
mass
was
2.27941 m from the left end. Now the
center of mass is 3.30882 m from the left
end.
But
the
center
of
mass
never
moved! This means the boat moved
3.30882 m − 2.27941 m =
1.02941 m
to the left as the man walked 1.02941 m
to the right.
003 (part 1 of 4) 10.0 points
Consider a rod of length L and mass m which
is pivoted at one end. An object with mass m
is attached to the free end of the rod.
m
◦
C
27
L
m
Determine the moment of inertia of the
system with respect to the pivot point.
The acceleration of gravity g = 9.8 m/s
2
.
Consider the mass at the end of the rod
to be a point particle.
1. I =
3
2
m L
2
2. None of these.
3. I =
4
3
m L
2
correct
4. I =
5
4
m L
2
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentCh9HW2 – florin – (56950)
2
5.
I =
13
12
m L
2
6.
I = L
2
7.
I =
5
3
m L
2
Explanation:
The moment of inertia of the rod
with respect to the pivot point is
I
ROD
=
1
3
m L
2
,
and the moment of inertia of the mass m with
respect to the pivot point is I
MASS
= m L
2
.
Thus the moment of inertia of the system is
I
=
I
ROD
+
I
M
=
1
3
m L
2
+ m L
2
=
4
3
m L
2
.
004 (part 2 of 4) 10.0 points
The length C in the figure represents
the location of the centerofmass of
the rod plus mass system.
Determine the position of the center
of mass from the pivot point.
005 (part 3 of 4) 10.0 points
The unit is released from rest in the
horizontal position.
What is the kinetic energy of the
unit when the rod momentarily has a
vertical orientation?
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '08
 Turner

Click to edit the document details