Answer:
(a) k = [tex]\frac{Mw^{2} }{6} (a^{2} +b^{2} )[/tex]
(b) τ = [tex]\frac{M}{3} (a^{2} +b^{2} )[/tex] ∝
Explanation:
The moment of parallel pipe rotating about it's axis is given by the formula;
I = [tex]\frac{M}{3} (a^{2} +b^{2} )[/tex] ---------------------------------1
(a) The kinetic energy of a parallel pipe is also given as;
k =[tex]\frac{1}{2} Iw^{2}[/tex] --------------------------------2
Putting equation 1 into equation 2, we have;
k = [tex]\frac{M}{6} (a^{2} +b^{2} )w^{2}[/tex]
k = [tex]\frac{Mw^{2} }{6} (a^{2} +b^{2} )[/tex]
(b) The angular momentum is given by the formula;
τ = Iw -----------------------3
Putting equation 1 into equation 3, we have
τ = [tex]\frac{Mw}{3} (a^{2} +b^{2} )[/tex]
But
τ = dτ/dt = [tex]\frac{M}{3} (a^{2} +b^{2} )\frac{dw}{dt}[/tex] ------------------4
where
dw/dt = angular acceleration =∝
Equation 4 becomes;
τ = [tex]\frac{M}{3} (a^{2} +b^{2} )[/tex] ∝
