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 7.00 kg , while the balls each have mass 0.300 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 D

Find the moment of inertia of this combination about an axis parallel to the bar and 0.500 m from it.

Explanation / Answer

A.) I' = Moment of inertia of ball 1 + Moment of inertia of ball 2 + MOI of rod through centre

I' = m (L/2)2 + m (L/2)2 + ML2 /12

I' = 0.3 x (2/2)2 + 0.3 x (2/2)2  + 7 x 22 /12 = 2.933333 kgm2

B.) I' =  Moment of inertia of ball 1 + Moment of inertia of ball 2 + MOI of rod through the end

I' = m (0)2 + mL2 + ML2/3

I' = 0 + 0.3 x 22 + 7 x 22 /3 = 10.533333 kgm2

D.) I' = Moment of inertia of ball 1 about new axis + Moment of inertia of ball 2 about new axis + MOI of rod about this new axis

I' = m r2 + mr2 + Mr2  

I' = (2m + M) r2   It is because in this case all the mass is at the same distance r = 0.5 from the new axis

I' = ( 2x 0.3 + 7 ) 0.52 = 1.9 kgm2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.