Question
A flywheel is making 180 r.p.m. and after 20 seconds it is running at 140 r.p.m. How many revolutions
will it make, and what time will elapse before it stops, if the retardation is uniform ?
will it make, and what time will elapse before it stops, if the retardation is uniform ?
Answers
V1=180rev/min * 6.28rad/rev * 1min/60s =
18.84 Rad/s.
V2 = 140/180 * 18.84 = 14.65 Rad/s.
a = (V2-V1)/t = (14.65-18.84)/20 = -0.209 Rad/s^2.
b. V3 = V1 + a*t = 0
t = -V1/a = -18.84/-0.209 = 90 s. To
stop.
a. V3^2 = V1^2 + 2a*d = 0
d = -V1^2/2a = -(18.84^2)/-0.418 = 849.2
Radians.
849.2Rad * 1rev/6.28Rad = 135.2 rev.
before it stops.
18.84 Rad/s.
V2 = 140/180 * 18.84 = 14.65 Rad/s.
a = (V2-V1)/t = (14.65-18.84)/20 = -0.209 Rad/s^2.
b. V3 = V1 + a*t = 0
t = -V1/a = -18.84/-0.209 = 90 s. To
stop.
a. V3^2 = V1^2 + 2a*d = 0
d = -V1^2/2a = -(18.84^2)/-0.418 = 849.2
Radians.
849.2Rad * 1rev/6.28Rad = 135.2 rev.
before it stops.
ans required
angular distance also can b found by s= (vi+vf) x t and the devided by 2
thanks
thanks
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