A wheel rotating at 1800 rpm slows down uniformly to 1200 rpm in 2.0 sec. Calculate:...

A wheel rotating at 1800 rpm slows down uniformly to 1200 rpm in 2.0 sec. Calculate:

(a) the angular acceleration (rev/s²) of the motor.

(b) the number of revolutions it makes in this time

Solutions

Expert Solution

(a) Let, the angular acceleration of the motor be .

Initial rate of rotation of its wheel was :

_i = 1800 rpm = 1800 revolutions / min = 1800 revolutions / 60 sec = 30 revolutions / sec.

Final rate of rotation of the wheel after time t = 2 sec :

_f = 1200 revolutions / min = 1200 revolutions / 60 sec = 20 revolutions / sec.

Hence, = ( _f - _i ) / t = ( 20 revolutions / sec - 30 revolutions / sec ) / 2 sec

or, the angular acceleration of the motor = = - 5 rev / s².

(b) Number of revolutions the wheel makes in this time :

n = _it + t² / 2 = 30 x 2 + ( - 5 ) x 2² / 2 = 50 revolutions.

genius_generous answered 1 year ago

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