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17. (a) The derivation of the acceleration is found in §114; Eq. 1113 gives
a
g
IM
R
com
com
=−
+
1
0
2
where the positive direction is upward. We use
I
com
gcm
=⋅
950
2
,
M
=120g,
R
0
= 0.320
cm and
g
= 980 cm/s
2
and obtain
()
(
)
2
22
com
2
2
980 cm/s


12.5 cm/s
13 cm/s .
1
950 g cm
120 g 0.32 cm
a
==
≈
+⋅
(b) Taking the coordinate origin at the initial position, Eq. 215 leads to
ya
t
com
com
=
1
2
2
.
Thus, we set
y
com
= – 120 cm, and find
com
2
com
2
120cm
2
4.38 s 4.4 s.
12.5 cm s
y
t
a
−
=
≈
−
(c) As it reaches the end of the string, its center of mass velocity is given by Eq. 211:
2
com
com
12.5 cm s
4.38s
54.8 cm s
va
t
−
=
−
,
so its linear speed then is approximately
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 Spring '08
 Any
 Physics, Acceleration

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