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MECHANICS. 30I
speed is in £ of a second reduced to a rate of 280 revolutions
per minute. What average force has the fly-wheel communi-
cated to the pitch-line of a 6-inch gear on the fly-wheel shaft ?
Solution
:
The mass of the fly-wheel = 644
= 20
32.2
The moment of rotation = {\y
2 f X 20 = 45
800 revolutions per minute = 5 revolutions per second.
Angular velocity = 5 X 6.2832 = 31.416
280 revolutions per minute — 42
/
3 revolutions per second.
Corresponding angular velocity = 4% X 6.2832 = 29.3216
6-inch diameter of gear = 3-inch radius = % foot.
F _ (31.416 — 29.3216) X 45
F = 2.0944 X 45 X 5 X 4 = 1884.96 pounds.
The kinetic energy in foot-pounds stored in the revolving
body may be obtained by the formula
:
Fa
2
X = kinetic energy
.
Decreasing the angular velocity to Fai, the stored-up
energy will also decrease to
w x
4
and the work done by the revolving body will be
(Fa
2 — Fal
2
) X -L
Example 1.
The moment of rotation in a fly-wheel is 1040 ; its angular
velocity is 5 feet per second. What is the stored-up energy in
the wheel ?
Solution :
Kinetic energy = 52
X 1040
= 13,000 foot-pounds.
Example 2.
At certain intervals, when machinery is started, the angular
velocity of this fly-wheel is reduced to 4^ feet per second.
How many foot-pounds of energy has the fly-wheel given up in
helping to drive the machinery ?
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