Answer:
(A) 0.63 J
(B) 0.15 m
Explanation:
length (L) = 0.75 m
mass (m) =0.42 kg
angular speed (ω) = 4 rad/s
To solve the questions (a) and (b) we first need to calculate the rotational inertia of the rod (I)
I = Ic + m[tex]h^{2}[/tex]
Ic is the rotational inertia of the rod about an axis passing trough its centre of mass and parallel to the rotational axis
h is the horizontal distance between the center of mass and the rotational axis of the rod
I = [tex](\frac{1}{12})(mL^{2} ) + m([tex]\frac{L}{2}[/tex])^{2}[/tex]
I = [tex](\frac{1}{12})(0.42 x 0.75^{2} ) + ( 0.42 x ([tex]\frac{0.75}{2}[/tex])^{2}[/tex])
I = 0.07875 kg.m^{2}
(A) rods kinetic energy = 0.5I[tex]ω^{2}[/tex]
= 0.5 x 0.07875 x [tex]4^{2}[/tex] = 0.63 J 0.15 m
(B) from the conservation of energy
initial kinetic energy + initial potential energy = final kinetic energy + final potential energy
Ki + Ui = Kf + Uf
at the maximum height velocity = 0 therefore final kinetic energy = 0
Ki + Ui = Uf
Ki = Uf - Ui
Ki = mg(H-h)
where (H-h) = rise in the center of mass
0.63 = 0.42 x 9.8 x (H-h)
(H-h) = 0.15 m
