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angular position. need work shown?
the angular position Q of a .36 m in diameter flywheel is given by
Q = (2.0 rad/s^3)t^3
i need to find Q in radians at t_1 = 2.0s and t_2 = 5.0s
and i need to find the distance the flywheel moves between t_1 and t_2
i know the radians after t_1 and t_2 are 16 and 250 respectively but the book does a poor job of explaining how it works. i dont understand how the equation works when substituting in for t
2 Answers
- oubaasLv 79 years agoFavorite Answer
Q1 = 2.0*t1^3 = 2*2^3 = 16 rad
Q2 = 2.0*t2^3 = 2*5^3 = 250 rad
ΔQ = Q2-Q1 = 250-16 = 234 rad
Each radians has a rectified lenght worth the radius of the circimference, i.e. half diameter (d/2); therefore distance D covered by flywheel is worth :
D = ΔQ*d/2 = 234*0.36/2 = 42,12 m
- FiremanLv 79 years ago
given :-Q = (2.0 rad/s^3)t^3
=>Q at t_1 = 2.0s :-
=>Q = 2 x (2)^3 = 16 rad
similerly Q at t_2 = 5.0s:-
=>Q = 2 x (5)^3 = 250 rad
=>âQ = Q(t_5) - Q(t_1)
=>âQ = 250 - 16 = 234 rad
BY s = r x âQ
=>s = 0.36 x 234
=>s = 84.24 m
