CHP. 9 CIRCULAR MOTION PRACTICE SOLUTIONS
1.
T
2.
T
3.
T
4.
F
5.
T
6.
A
7.
C
8.
D
9.
B
10. D
11. C
12. C
13. C
14. The whole pole rotates about its base as it falls at some rotational velocity. So the
far end is traveling faster than the center….twice faster. Since the center hits at
10 m/s, the far end hits at 20 m/s.
15.
This problem is the same as problem 13, above. The answer is given at the
beginning of the last paragraph on page 132 of the text.
Simulated gravity’s effect
rests on the fictitious centrifugal force which is an “outward” feel from the
centripetal force on the rotating wall on which we’re “standing”. Now this force is
directly proportional as :
F
c
~ (m v
2
)/r , but v is directly proportional as v ~ r ω.
So if we substitute in for v
2
in the expression for F
c,
we get:
F
c
~ (m v
2
)/r
~
m (r ω)
2
/r
~
m r
2
ω
2
/r
~
m r ω
2
The acceleration (simulated gravity) felt by a person of mass m is then directly
proportional as:
a = F
c
/m ~ r ω
2
so
if
you go from 200 meters from the axis
of rotation to 100 meters from that axis, the aceleration you feel will halve. While
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 Spring '08
 Staff
 Physics, Circular Motion, Rotation, 20 m/s, FC, 10 m/s, 100 meters, 200 meters

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