homework 11 – – Due: Apr 12 2007, 4:00 am
1
Question 1, chap 13, sect 2.
part 1 of 2
8 points
A spool of wire
of mass
M
= 5
.
4 kg
and
radius
R
=
0
.
64
m
is
unwound
under
a
constant
wire
tension
F
=
4
.
3
N.
Assume
the
spool
is
a
uniform
solid
cylinder
that
rolls
without
slipping.
R
F
M
Calculate the friction force
f
on the bottom
of the spool. Take the rightward direction to
be positive.
Correct answer:
1
.
43333
N (tolerance
±
1
%).
Explanation:
By Newton’s Second Law, the spool’s cen
ter of mass (which is in the middle of the
spool’s axis) accelerates according to
Ma
=
F
+
f
(1)
where
F
is the wire tension,
f
is the friction
force at the bottom of the spool, and the pos
itive direction for both forces is towards the
right.
At the same time, the spool’s angu
lar acceleration obtains from the net torque
in the centerofmass frame, thus
Iα
=
F
×
R

f
×
R
(2)
where
I
≡
I
cm
=
MR
2
2
.
(3)
Finally, since the spool rolls without slipping,
α
=
a
R
.
(4)
Substituting eqs. (3) and (4) into eq. (2) we
find
MRa
2
=
FR

fR
=
⇒
Ma
= 2
F

2
f
(5)
Comparing this formula with eq. (1), we see
that
2
F

2
f
=
F
+
f
(6)
and therefore
f
=
F
3
= 1
.
43333 N
.
(7)
Question 2, chap 13, sect 2.
part 2 of 2
8 points
Calculate the acceleration of the spool’s
center of mass.
Correct answer: 1
.
06173 m
/
s
2
(tolerance
±
1
%).
Explanation:
According to eqs. (1) and (7)
Ma
=
F
+
f
=
4
3
F
(8)
and therefore
a
=
4
F
3
M
= 1
.
06173 m
/
s
2
.
Question 3, chap 13, sect 2.
part 1 of 1
8 points
Consider a basketball rolling (without slip
ping) down an inclined plane. The basketball
is basically a thin spherical shell, hence its
moment of inertia
I
=
2
3
M R
2
.
The acceleration of gravity is 9
.
8 m
/
s
2
.
Given the angle
θ
= 32
.
6
◦
between the
incline and the horizontal, calculate the bas
ketball’s center’s acceleration.
Hint:
You don’t need the values of the
basketball’s mass or radius.
Correct answer: 3
.
16797 m
/
s
2
(tolerance
±
1
%).
Explanation:
Rolling motion of the basketball comprises
the linear motion of the basketball’s center
of mass (which is of course at the geometric
center of the ball) and the rotation around the
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homework 11 – – Due: Apr 12 2007, 4:00 am
2
axis going through that center. The rotation
governed by the equation
I α
=
τ
net
,
where the right hand side is the net torque
of all the forces acting on the basketball.
There are three such forces — the basket
ball’s weight
vector
W
=
M vectorg ,
the normal force
vector
N
and the friction force
vector
f
— but the
vector
W
and the
vector
N
act along lines going through the ball’s cen
ter and hence have zero lever arms.
On the
other hand, the lever arm of the friction force
is the ball’s radius
R
, hence
τ
net
=
W
×
0 +
N
×
0 +
f
×
R ,
and consequently
I α
=
f R .
Now consider the linear motion of the bas
ketball’s center along the incline; by Newton’s
Second Law,
M a
=
F
net
along
=
M g
sin
θ

f .
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 Spring '08
 Turner
 Force, Friction, Mass, Work, Moment Of Inertia, Correct Answer, Eqs

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