8.28 Problem = 1 2 2
=
=2 =
=2
1 2 2 2
=
=
=
+
1 2 2
=2
=
1 2 2
2
+
1 2 2
=
1 8
+
1 4
=
3 8
Conservation of energy dictates that: 1 2 1 2 = 3 8 6 8 3 8
+
+
=
+
2 1 2
+
2 6 8 = +
Simplifying: + = + + 2 2
11 16 =
=
3 2
Solving for .
Assuming constant acceleration: =2 = 12 11
Problem 8.32
= =2 = = 10 1 3
= 1- 4
= =2
- =2 -1 - 5
Substituting in =
Yields: =
3 10
Problem 8.41 = = 1 2 +
=
1 2
1 2 3 4
=
1 2 1 2 2 3
+
1 1 2 2
=
3 4
- =
=
-
3 2 = 1 2
=
-2 3
=
-
=
=
=
+
=
3 2
3 2
=
+
=0
=
2 3
Problem 8.42 Left end is pivot point. = + = = - -2 + + = - =0 = + = = + +
1 + +
+ 2 +
+
-2 +
-2
The maximum will be when this is equal to zero:
= 2 - 1
Problem 8.61 Conservation of angular momentum: = = = 2
=2
2 2 (1)
= = 1 2
1 2
=
Conservation of linear momentum: x-direction: =2 where (2) is the x component of the bar's velocity.
= 0=
y-direction: -2 where (3) is the y component of th...