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 ?

5 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.
ans required
angular distance also can b found by s= (vi+vf) x t and the devided by 2
thanks
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