A uniform bar has two small balls glued to its ends. The bar is 2.00 m long and with mass 4.00 kg, while the balls each have mass 0.500 kg and can be treated as point masses.
A) Find the moment of inertia of this combination about an axis perpendicular to the bar through its center.
B) Find the moment of inertia of this combination about an axis perpendicular to the bar through one of the balls.
C) Find the moment of inertia of this combination about an axis parallel to the bar through both balls.
D) Find the moment of inertia of this combination about an axis parallel to the bar and 0.500 m from it.
Find the moment of inertia of this combination about an axis perpendicular to the bar through its center.?
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Suppose M is mass of bar and L is its length, then moment of inertia of the bar about an axis perpendicular to the bar through its center=Ic=ML^2/12
M=4.00 kg
L=2.00m
Ic=4*2^2/12= 4/3 kgm^2
If m is mass of ball and r is distance from the axis of rotation,(r=L/2=1.00m)
moment of inertia of each ball =Ib = mr^2=0.5*1*1=0.5 kgm^2
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A) The moment of inertia of the combination about an axis perpendicular to the bar through its center=Ic+Ib=1.33+2*0.5=2.33 kgm^2
The moment of inertia of the combination about an axis perpendicular to the bar through its center is 2.33 kgm^2
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B) Moment of inertia of the bar about an axis perpendicular to the bar through its end=Ie=ML^2/3
Ie=4*2*2/3=16/3=5.333 kgm^2
moment of inertia of the ball through which axis passes is zero
moment of inertia of the other ball about an axis through the other end of bar=Ib=mr^2=mL^2
moment of inertia of the other ball =Ib=0.5*2^2= 2 kgm^2
the moment of inertia of this combination about an axis perpendicular to the bar through one of the balls=I1=5.33+2=7.33 kgm^2
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C) Moment of inertia of the combination about an axis parallel to bar through both the balls is zero because center of mass of bar and balls lie on the axis of rotation
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D) the moment of inertia ,Ipar, of this combination about an axis parallel to the bar and 0.500 m from it is found using parallel axis theorem
Ipar= I+M!h^2
where I is moment of inertia of the system about axis passing through the bara nd both the balls and M! =sum of masses of the two balls =m+m= o.5+0.5=1.00 kg
Ipar=zero+[0.5+0.5]0.25=0.25kgm^2
the moment of inertia of this combination about an axis parallel to the bar and 0.500 m from it is 0.25 kgm^2
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shark teeth
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for part D, the sum of the masses should also incorporate the mass of the bar, 4kg. Therefore you would get [4+.5+.5].25=1.25kgm^2. This is the correct answer.
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