Question
A wheel rotating initially at an angular velocity of 1.0 rad/s, accelerates over a 10 S interval at a rate of 0.1 rad/s^2. What angular displacement does the wheel undergo in this 10 S interval, in terms of revolutions?
Answers
Henry
D = Vo*t + 0.5a*t^2.
D = 1*10 + 0.05*10^2 = 15 rad.
D = 15rad/(6.28rad/rev.) = 2.39 rev.
D = 1*10 + 0.05*10^2 = 15 rad.
D = 15rad/(6.28rad/rev.) = 2.39 rev.