Respuesta :
Here the maximum angular displacement of the sign board is given 25 degree
mass of the sign is 2.40 kg and its vertical side is 55 cm
Now the moment of inertia of the board about the hinge point is given as
[tex]I = \frac{mL^2}{3}[/tex]
[tex]I = \frac{2.40\times 0.55^2}{3} = 0.242 kg m^2[/tex]
now by energy conservation
[tex]\frac{mgL}{2}(1 - cos\theta) = \frac{I\omega^2}{2}[/tex]
[tex](2.40)(9.8)(0.55)(1 - cos25) = 0.242(\omega)^2[/tex]
[tex]\omega = 2.24 rad/s[/tex]
so angular speed just before the impact is 2.24 rad/s
Part b)
now a mass of 450 g travels in opposite direction and hit at the lower end
so here we will use angular momentum conservation
[tex]I_1\omega_1 - mvL = (I + mL^2)\omega_2[/tex]
[tex]0.242(2.24) - (0.45\times 1.60 \times 0.55) = (0.242 + (0.45\times 0.55^2)\omega_2[/tex]
[tex]0.542 - 0.396 = (0.378)\omega_2[/tex]
[tex]\omega_2 = 0.386 rad/s[/tex]
Part c)
Again we can use energy conservation
[tex](\frac{m_1gL}{2}+ m_2gL)(1 - cos\theta) = \frac{(I_1 + I_2)\omega^2}{2}[/tex]
[tex](\frac{(2.40)(9.8)(0.55)}{2} + 0.450\times 9.8 \times 0.55)(1 - cos\theta) = \frac{(0.242+ 0.136)}{2}(0.386)^2[/tex]
[tex]8.89( 1- cos\theta) = 0.028[/tex]
[tex]cos\theta = 0.997[/tex]
[tex]\theta = 4.56 degree[/tex]
