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A uniform bar has two small balls glued to its ends. The bar is 2.10 m long and

ID: 1445928 • Letter: A

Question

A uniform bar has two small balls glued to its ends. The bar is 2.10 m long and with mass 4.00 kg , while the balls each have mass 0.600 kg and can be treated as point masses. Part A Find the moment of inertia of this combination about an axis perpendicular to the bar through its center. Part B Find the moment of inertia of this combination about an axis perpendicular to the bar through one of the balls. Part C Find the moment of inertia of this combination about an axis parallel to the bar through both balls. I=______kg*m^2 for a,b,c

Explanation / Answer

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=2.10m
Ic=4*2.1^2/12= 1.47 kgm^2

If m is mass of ball and r is distance from the axis of rotation,(r=L/2=1.05m)

moment of inertia of each ball =Ib = mr^2=0.6*1.05*1.05=0.6 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=1.47+2*0.6=2.67 kgm^2

The moment of inertia of the combination about an axis perpendicular to the bar through its center is 2.67 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*2.1*2.1/3=16/3=5.88 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=0.6*2.1^2= 2.646 kgm^2

the moment of inertia of this combination about an axis perpendicular to the bar through one of the balls=I1=5.88+2.646=8.526 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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