Asked by Rupsa
If a wheel revolving at 1800 r.p.m. slows down uniformly to 1200 r.p.m. in 25s, calculate the angular acceleration of the wheel and the number of revolutions it makes in this time.
Answers
Answered by
Anonymous
1800 rev/min * 2 pi rad/rev * 1 min/60 s = Omegai
1200 similarly ---> Omegaf
d omega = (Omegaf-omegai)
dt = 25 s
so
d Omega/dt = alpha = d omega/dt
average angular velocity
= (Omagaf+Omegai)/2
total radians = average vel * 25 sec
total revs = total radians/2pi
1200 similarly ---> Omegaf
d omega = (Omegaf-omegai)
dt = 25 s
so
d Omega/dt = alpha = d omega/dt
average angular velocity
= (Omagaf+Omegai)/2
total radians = average vel * 25 sec
total revs = total radians/2pi
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