=
k , with = 0
r = ber , so ( r ) = b 2 er
v = ve , so v = ver
From eqn 5.2.14,
a = a + 2 v + ( r ) and putting in terms from above
v 2
2 v b 2
ar =
b
For no slipping F s mg , so a s g
v 2
+ 2 v + b 2 s g
b
2
vm + 2 bvm + b 2 2 b s g = 0
vm = b 2b 2 b 2 2 + b s g
Since v was defined positive, the +square root is used.
vm = b + b s g
(b)
v ve =
v = + ver
2
v
+ 2 v b 2
ar =
b
2
v
2 v + b 2 s g
b
vm = b + b s g
2
5.9
V
V
i
As in Example 5.2.2, = and A =
j
For the point at the front of the wheel:
2
V
r =
j and v = V k
b
=0
V
Vb
j
i
r = k b =
( r ) =
v =
(
)
2
V Vb V b
k
i =
j
(
)
V
k V k = 0
2
2
2
V V
V b
a = r + ( r ) + A =
i +
+ 2 j
b
5