A Motor revolving at 1200 rpm slows down uniformly to 900 rpm in 2 sec. Calculate the angular acceleration of the motor and number of revolution it makes during this time.

Explanation

Given ` omega_0 = 1200 "rpm" = (1200xx 2pi) /( 60 ) rad s^(-1)`
i.e., `omega_0 = 40 pi rad s^(-1) `
` omega_f = omega = 900 ` rpm
` = (900 xx 2pi)/(60) = rad s^(-1)`
` omega = 30 pi rad s^(-1)`
wkt ` omega = omega_0 + alpha t `
i.e., `alpha = ( omega - omega_0)/( t)`
i.e., ` alpha = ((30 - 40) pi)/(2)`
` alpha = - 2pi rad s^(-2)`
Hence angular acceleration ` = - 5pi rad s^(-2)`
wkt ` theta = w_0 t + (1/2) alphat^2`
i.e., ` theta = (40 pi ) 2 + (1/2) (-5pi) (2)^2`
i.e., ` theta = 80 pi - 10 pi = 70 pi rad`
No. of revolution made ` = (theta)/(2pi) = (70 pi)/(2pi)`
`N = 35 `