The moment of inertia is [tex]24.8 kg m^2[/tex]
Explanation:
The total moment of inertia of the system is the sum of the moment of inertia of the rod + the moment of inertia of the two balls.
The moment of inertia of the rod about its centre is given by
[tex]I_r = \frac{1}{12}ML^2[/tex]
where
M = 24 kg is the mass of the rod
L = 0.96 m is the length of the rod
Substituting,
[tex]I_r = \frac{1}{12}(24)(0.96)^2=1.84 kg m^2[/tex]
The moment of inertia of one ball is given by
[tex]I_b = mr^2[/tex]
where
m = 50 kg is the mass of the ball
[tex]r=\frac{L}{2}=\frac{0.96}{2}=0.48 m[/tex] is the distance of each ball from the axis of rotation
So we have
[tex]I_b = (50)(0.48)^2=11.5 kg m^2[/tex]
Therefore, the total moment of inertia of the system is
[tex]I=I_r + 2I_b = 1.84+ 2(11.5)=24.8 kg m^2[/tex]
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