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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.

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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