rotatory motion
Asked by | 27th Jun, 2009, 07:24: PM
(i) omega0 = 1200 rpm = 40 pi rad / s;
omega = 3120 rpm = 104 pi rad / s
using omega = omega0 + alpha x t
104 pi = 40 pi + alpha x 16
alpha = 66 pi / 16 = 33 pi / 8 rad / s2
(ii) theta = omega0 x t + (1/2) x alpha x t2
theta = 40 pi x 16 + (1/2) x (33 pi / 8) x 16 x 16 = 640 pi + 528 pi = 1168 pi
So number of revolutions = 1168 pi / 2 pi = 584
Answered by | 7th Jul, 2009, 02:46: PM
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