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Attempt any SIX : The frequency of revolution of a particle performing circular motion changes from 60 r.p.m to 180 r.p.m in 20 seconds. Calculate the angular acceleration of the particle. |
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Answer» Given : `n_(1) = 60 r.p.m. = (60)/(60)` r.p.s = 1 Hz `n_(2) = 180 `r.p.m. `= (180)/(60)` r.p.s. `= 3 Hz` `t = 20` sec. `alpha = (omega_(2) - omega_(1))/(t)` `alpha = (2pi(n_(2) - n_(1)))/(t)` `alpha = (2pi(3-1))/(20)` `alpha = (2pi)/(10) = (2 xx 3.142)/(10)` `alpha = 0.6284 "rad"//s^(2)` |
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