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Oct. 4, 2010
PHYS 2101
HW#6 WileyPlus Problem Solutions
1
HW #6 Solutions – or at least most of them…
(USING BOOK VALUES 9
th
Edition in problems; problem # indicated by [])
1) [113] By Eq. 1052, the work required to stop the hoop is the negative of the initial kinetic energy of
the hoop. The initial kinetic energy is
(Eq. 115), where
I = mR
2
is its rotational inertia
about the center of mass,
m
= 140 kg, and
v
= 0.150 m/s is the speed of its center of mass. Equation 112
relates the angular speed to the speed of the center of mass:
!
=
v
/
R
. Thus,
which implies that the work required is
.
2) [117] (a) We find its angular speed as it leaves the roof using conservation of energy. Its initial kinetic
energy is
K
i
= 0 and its initial potential energy is
U
i
= Mgh
where
(we are using the
edge of the roof as our reference level for computing
U
). Its final kinetic energy (as it leaves the roof) is
(Eq. 115)
.
Here we use
v
to denote the speed of its center of mass and
is its angular speed — at the moment it
leaves the roof. Since (up to that moment) the ball rolls without sliding we can set
v
=
R
=
v
where
R
=
0.10 m. Using
(Table 102(c)), conservation of energy leads to
The mass
M
cancels from the equation, and we obtain
=
1
R
4
3
gh
=
1
0.10 m
4
3
2
( )
3.0 m
( )
=
.
(b) Now this becomes a projectile motion of the type examined in Chapter 4. We put the origin at the
position of the center of mass when the ball leaves the track (the
“initial”
position for this part of the
problem) and take +
x
leftward and +
y
downward. The result of part (a) implies
v
0
=
R
= 6.3 m/s, and we
see from the figure that (with these positive direction choices) its components are
The projectile motion equations become
We first find the time when
y
=
H
= 5.0 m from the second equation (using the quadratic formula,
choosing the positive root):
Then we substitute this into the
x
equation and obtain
x
=
5.4 m s
( )
0.74 s
( ) =
4.0 m.
3) [1110] . From
(Table 102(g)) we find
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View Full DocumentOct. 4, 2010
PHYS 2101
HW#6 WileyPlus Problem Solutions
2
It also follows from the rotational inertia expression that
. Furthermore, it rolls without
slipping,
v
com
=
R
!
, and we find
(a) Simplifying the above ratio, we find
K
rot
/
K
= 0.4. Thus, 40% of the kinetic energy is rotational, or
K
rot
= (0.4)(20 J) = 8.0
J.
(b) From
(and using the above result for
M
) we find
=
1
0.15 m
3 8.0 J
( )
2.7 kg
=
which leads to
v
com
= (0.15 m)(20 rad/s) = 3.0 m/s.
(c) We note that the inclined distance of 1.0 m corresponds to a height
h
= 1.0 sin 30° = 0.50 m.
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