Answer:
the length of the boomerang must be = 36.65 cm
Explanation:
Given that:
diameter of both combination of a thin outer hoop and a uniform disk [tex]D _d[/tex] = 0.273 m
The mass of the hoop part is [tex]m _h[/tex] = 0.120 kg
The mass of the disk part is [tex]m _d[/tex] = 0.050 kg
The total mass of the boomerang is to be [tex]m _b[/tex] = 0.255 kg,
what must be the length [tex]L _b[/tex] of the boomerang = ???
The moment of inertia of the sport disk is :
[tex]{I = I_1} + I_2}[/tex]
Let [tex]I_1}[/tex] be the moment of inertia of the hoop :
SO; [tex]I_1}[/tex] [tex]= m_h*r^2[/tex]
[tex]I_1}[/tex] [tex]= 0.120*(\frac{0.273}{2})^2[/tex]
[tex]I_1}[/tex] [tex]= 0.120*(0.1365)^2[/tex]
[tex]I_1}[/tex] = 0.00223587 kgm²
Let [tex]I_2}[/tex] be the moment of inertia of the disc
SO; [tex]I_2}[/tex] [tex]= \frac{1}{2}*m_d*r^2[/tex]
[tex]I_2}[/tex] [tex]= \frac{1}{2}*0.050*(\frac{0.273}{2})^2[/tex]
[tex]I_2}[/tex] [tex]= 0.5*0.050*(0.1365)^2[/tex]
[tex]I_2}[/tex] = = 0.0004658 kgm²
NOW;
[tex]{I = I_1} + I_2}[/tex]
I = (0.00223587 + 0.0004658) kgm²
I = 0.00270167 kgm²
However; given that I is the moment of the boomerang
[tex]{I = I_1} + I_2}[/tex]
[tex]I_1} = I_2}[/tex] (i.e the moment of inertia in each rod )
[tex]I = 2*I_1[/tex]
[tex]0.00270167 = 2*\frac{1}{2}*\frac {0.255}{2}*lb^2\\\\\\0.00270167 = 0.02125 *lb^2\\\\lb^2 = \frac{0.00270167}{0.02125}\\\\ \\lb^2 = 0.1271\\\\lb = \sqrt{0.1271}\\\\lb = 0.3565 \ \ m\\\\lb = 36.65 \ \ cm[/tex]
Therefore; the length of the boomerang must be = 36.65 cm
