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PHY 2053
Spring 2011
Recitation # 6
Chapter # 4
Feb 21 – Feb 25, 2011
10.
The large blade of a helicopter is rotating in a horizontal circle. The length of the blade is 6.7 m,
measured from its tip to the center of the circle. Find the ratio of the centripetal acceleration at the
end of the blade to that which exists at a point located 3.0 m from the center of the circle.
10.
REASONING
The magnitude of the centripetal acceleration of any point on the helicopter
blade is given by Equation 5.2,
a
C
=
v
2
/
r
, where
r
is the radius of the circle on which that
point moves. From Equation 5.1:
v
r
T
=
2
π
/
. Combining these two expressions, we obtain
a
r
T
C
=
4
2
2
All points on the blade move with the same period
T
.
SOLUTION
The ratio of the centripetal acceleration at the end of the blade (point 1) to that
which exists at a point located 3.0 m from the center of the circle (point 2) is
a
a
r
T
r
T
r
r
C1
C2
6.7 m
3.0 m
2.2
=
=
=
=
4
4
2
1
2
2
2
2
1
2
/
/
15.
A 0.015kg ball is shot from the plunger of a pinball machine. Because of a centripetal force of
0.028 N, the ball follows a circular arc with a radius of 0.25 m. What is the speed of the ball?
15.
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 Spring '08
 Vellaris
 Physics

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