A motor revolving at 1200 RPM slows down uniformly to 900 RPM in 2 seconds. Calculate the angular acceleration of the motor and the number of revolutions it takes during this time.
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this is just example
Angular speed of a wheel is increased from 1200 rpm to 3120 rpm in 16 seconds. (i) What is the angular acceleration (Assume the acceleration is uniform)? (ii) How many revolution does the wheel make during this time?
initial angular velocity, w1 = 1200 rpm = 1200*2pi/60 = 40 pi rad/s
Final angular velocity, w2 = 3120 rpm = 3120*2pi / 60 = 104 pi rad/s
Time t = 16 s
(i)Angular acceleration, a = (104-40)/16 = 64pi/16 = 4 pi rad/s^2
(ii) Angle, theta = 40pi*16 + ½*4pi * 256 = 1152pi rad
Thus, number of revolutions, n= 1152pi/2pi revolutions
= 576 rev
I hope this will help you
if not then comment me
Angular speed of a wheel is increased from 1200 rpm to 3120 rpm in 16 seconds. (i) What is the angular acceleration (Assume the acceleration is uniform)? (ii) How many revolution does the wheel make during this time?
initial angular velocity, w1 = 1200 rpm = 1200*2pi/60 = 40 pi rad/s
Final angular velocity, w2 = 3120 rpm = 3120*2pi / 60 = 104 pi rad/s
Time t = 16 s
(i)Angular acceleration, a = (104-40)/16 = 64pi/16 = 4 pi rad/s^2
(ii) Angle, theta = 40pi*16 + ½*4pi * 256 = 1152pi rad
Thus, number of revolutions, n= 1152pi/2pi revolutions
= 576 rev
I hope this will help you
if not then comment me
Ramyapanda:
the solution is elegant but why did u take t=16s given data is something else
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