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
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '08
 Any
 Physics, Acceleration

Click to edit the document details