Answer:
0.4112 m
Explanation:
The mass of the 1st ball = 3.6 kg
The height of the 1st ball =3.5 m
The mass of the 2nd ball = 3.6 kg
Mass of the bar M = 9.9 kg
Length of the bar L = 4.2 m
The velocity of the ball when it dropped from the height is calculated by using the formula:
[tex]\dfrac{1}{2}mv_1^2 = mgh_1 \\ \\ v = \sqrt{2gh_1} \\ \\ v =\sqrt{2\times9.8 \times 3.5} \\ \\ v = 8.283 \ m/s\\\\[/tex]
Provided that the bar is pivoted at the center and the ball is placed at the two ends, the moment of inertia for the bar is:
[tex]I = \dfrac{1}{12}ML^2 + m_1 (\dfrac{L}{2})^2 + m_2(\dfrac{L}{2})^2 \\ \\ =\dfrac{1}{12}(9.9kg)(4.2m)^2 + [3.6 kg+3.6kg](\dfrac{4.2}{2 \ m})^2 \\ \\ = 46.305 \ kg.m^2[/tex]
The angular momentum of the system due to the ball can be determined by using the formula:
L = mvr
L = (3.6 kg) (8.283 m/s) (2.1 m)
L = 62.61948 kg. m²
Now, Using the law of conservation:
[tex]L_i = L_f \\ \\ 62.61948 \ kg.m^2/s = I \omega \\ \\[/tex]
[tex]\omega = \dfrac{62.6198 \ kg.m^2/s}{46.305 \ kg.m^2}[/tex]
[tex]\omega =1.352 \ rad/s[/tex]
The linear angular velocity is deduced to be:
[tex]v = r \omega \\ \\ v = (2.1 \ m) ( 1.352 \ rad/s) = 2.839 \ m/s[/tex]
∴
the height raised by the second ball is:
[tex]h_2 = \dfrac{v^2}{2g} \\ \\ h_2 = \dfrac{(2.839)^2}{2(9.8 \ m/s^2)} \\ \\ h_2 =0.4112 \ m[/tex]
