A uniform bar has two small balls glued to its ends. The bar is 2.00 m long and

Question

A uniform bar has two small balls glued to its ends. The bar is 2.00 m long and has mass 4.50 kg , while the balls each have mass 0.300 kg and can be treated as point masses.Part B
Find the moment of inertia of this combination about an axis perpendicular to the bar through one of the balls.
Express your answer with the appropriate units.

I =

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Part C
Find the moment of inertia of this combination about an axis parallel to the bar through both balls.
Express your answer with the appropriate units.

I =

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Part D
Find the moment of inertia of this combination about an axis parallel to the bar and 0.500 m from it.
Express your answer with the appropriate units.

I =
3.375kg•m2

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Explanation / Answer

B) for the bar
Ibar=M*L^2/3=4.50*4/3=18/3 =6 kg-m^2
for the point masses Iball= mL^2=0.3*4=1.2 kg-m^2

I of combination = 6+1.2 =7.2 kg-m^2

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

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+Mh^2
where I is moment of inertia of the system about axis passing through the bar and both the balls and M =sum of masses of the two balls =m+m= o.3+0.3=0.6 kg
Ipar=zero+[0.3+0.3]0.25=0.15kgm^2
the moment of inertia of this combination about an axis parallel to the bar and 0.500 m from it is 0.15 kgm^2

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