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Dynamics
827
Problems 7.25 and 7.26
A bowling ball is thrown onto a lane with a backspin
!
0
and forward velocity
v
0
. The mass of the ball is
m
, its radius is
r
, its radius of gyration is
k
G
, and
the coefFcient of kinetic friction between the ball and the lane is
±
k
. Assume
the mass center
G
is at the geometric center.
Problem 7.25
±ind the acceleration of
G
and the ball’s angular acceleration
while the ball is slipping.
Problem 7.26
±or a
14
lb ball with
r
D
4:25
in
:
,
k
G
D
2:6
in,
!
0
D
10
rad
=
s,
and
v
0
D
17
mph, determine the time it takes for the ball to start rolling without
slip and its speed when it does so. In addition, determine the distance it travels
before it starts rolling without slip. Use
±
k
D
0:10
.
Solution to 7.25
Based on the ±BD shown, the NewtonEuler equations are
X
F
x
W
±
F
D
ma
Gx
;
(1)
X
F
y
W
N
±
mg
D
ma
Gy
;
(2)
X
M
G
W
±
Fr
D
I
G
˛
b
;
(3)
where
˛
b
is the angular velocity of the ball and where
I
G
D
mk
2
G
:
(4)
Since the ball slides (at least at Frst) over the lane, the force law for this problem is
F
D
±
k
N:
(5)
The kinematic equations consist only of the following relation:
a
Gy
D
0:
(6)
Substituting the force law, the kinematic equations, and the expression for the mass moment of inertia into
the NewtonEuler equations gives three equations in the three unknowns
N
,
a
, and
˛
b
whose solution is
N
D
mg
a
D ±
±
k
g;
and
˛
b
gr±
k
k
2
G
:
Hence, the Fnal answer is
a
±
k
g
and
˛
b
k
k
2
G
:
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Solutions Manual
Solution to 7.26
In this problem, the governing equations are identical to those of Problem 7.25. Therefore, we have that the
solution of this problem in terms of acceleration of the mass center and angular acceleration of the ball are
the same as those of Problem 7.25.
To determine when the ball starts slipping, we need to determine the velocity of the point on the ball that
is in contact with the ground as a function of time and then determine how long it takes for that velocity to
become equal to zero. Observing that the computed accelerations are constant, applying constant acceleration
equations, we have that the velocity of the mass center
G
as a function of time is
E
v
G
D
.v
0
±
±
k
gt/
O
{:
(7)
Again using constant acceleration equations, we have that the angular velocity of the ball is
E
!
b
D
!
0
±
±
k
rg
k
2
G
t
!
O
k:
(8)
Denoting the point on the ball that is in contact with the ground by
D
and observing that the position of
D
relative to
G
is
E
r
D=G
D ±
r
O

, we have
E
v
D
DE
v
G
CE
!
b
² E
r
D=G
D
"
v
0
C
r!
0
±
gt±
k
1
C
r
2
k
2
G
!#
O
(9)
The velocity of point
D
is the velocity with which the ball slips relative to the Foor. The time required to
achieve rolling without slip is therefore the time it takes for
E
v
D
D
E
0
, i.e.,
E
v
D
D
E
0
)
v
0
C
0
±
gt
r
±
k
1
C
r
2
k
2
G
!
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This note was uploaded on 12/04/2010 for the course ECE 134 taught by Professor York during the Fall '08 term at UCSB.
 Fall '08
 York

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