10. (6 pts.)
Note the following three cases. In which case will the magnitude of the normal force
on the block be equal to (
Mg
+
F
sin
θ
)?
A.
Case 1 only
B.
Case 2 only
C.
Both cases 1 and 2
D.
Both cases 2 and 3
E.
Cases 1, 2, and 3
11. (6 pts.)
The figure gives the angular momentum L of a wheel versus time t. Which of the
following is true for the
magnitudes
of the torques in the intervals shown?
B
C
L
t
B
C
A
τ
τ
τ
τ
>
>
=
C
B
A
D
τ
τ
τ
τ
>
>
>
A
A
A
12. (6 pts.):
A bullet is aimed at a target on the wall a distance
L
away from the firing position.
Because of gravity, the bullet strikes the wall a distance
Δ
y
below the mark as suggested in the
figure.
Note: The drawing is not to scale.
If the distance
L
were half as large, and the bullet had
the same initial velocity, how would
Δ
y
be affected?
A
B
C
L
t
D
2
v
gr
=
v
gr
=
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D
B
C
A
τ
τ
τ
τ
>
>
=
C
B
A
D
τ
τ
τ
τ
>
>
>
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Problem 1 (25 pts.):
A bowling ball of mass
M
and radius
R
(
treated as a uniform solid sphere
)
is thrown along a
level
floor with kinetic energy
K
0
. It is thrown so that initially (at
t
= 0) it slides
with a linear speed
ν
0
but does not rotate. As
the ball slides, kinetic friction causes it to
begin spinning. Eventually the ball begins to
roll without slipping along the floor, with
linear speed
ν
and angular speed
ω
, as shown
below. The coefficient of kinetic friction
between the ball and the floor is
μ
k
.
a) (
15 pts.
) At what time
t > 0
does the bowling ball begin rolling smoothly along the floor?
What is the linear speed
ν
at that time?
Express your answers in terms of
ν
0
, g, and μ
k
, as needed.
I
=
2
5
MR
2
(solid sphere)
Ma
=
−
f
=
−
μ
k
Mg
⇒
a
=
−
μ
k
g
(COM deceleration)
I
α
=
−
fR
=
−
μ
k
MgR
⇒
α
R
=
5
2
(
−
μ
k
g
) (acceleration of rotation)
v
=
v
0
+
at
=
v
0
−
μ
k
gt
(linear motion of COM)
ω
R
=
α
Rt
=
5
2
(
−
μ
k
g
)
t
(Rotation motion)
Rolling condition:
v
=
−
ω
R
at
t
=
t
r
⇒
v
0
+
(
−
μ
k
g
)
t
r
=
−
5
2
(
−
μ
k
g
)
t
r
=
5
2
μ
k
gt
r
v
0
=
7
2
μ
k
gt
r
⇒
t
r
=
2
v
0
7
μ
k
g
b) (
10 pts.
) What is the change
Δ
K
in the bowling ball’s kinetic energy during this entire process
(
i.e
., from sliding motion to rolling without slipping)?
Express your answer in terms of
K
0
.
K
0
=
1
2
Mv
0
2
(initial K.E) and
K
r
=
1
2
Mv
r
2
+
1
2
I
ω
r
2
(rolling K.E.)
K
r
=
1
2
Mv
r
2
+
1
2
I
ω
r
2
=
1
2
Mv
r
2
(1
+
I
MR
2
)
=
1
2
Mv
r
2
(
7
5
)
But
v
r
=
v
0
+
at
r
=
v
0
−
μ
k
gt
r
=
v
0
−
(
2
7
)
v
0
=
5
7
v
0
[from the result in (a)]
K
r
=
1
2
Mv
r
2
(
7
5
) =
1
2
M
5
7
v
0
!
"
#
$
%
&
2
(
7
5
)
=
5
7
K
0
Δ
K
=
K
r
−
K
0
=
−
2
7
K
0
v
o
ω
0
= 0
f
x
r
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