ok seriously i\'ve done this problem and i know concept everything! for this one

Question

ok seriously i've done this problem and i know concept everything!

for this one.. i sphere its 2/5mr^2..

so when i do 2/5mr^2( sphere ball left)+2/5mr^2(sphere ball right ) +1/12 ml^2 (rod)

doesnt give me 20 kg

A part of a mobile suspended from the ceiling of an airport terminal building consists of two metal spheres, each with mass 2.0 kg. connected by a uniform metal rod with mass 3.0 kg and length s = 4.0 m The assembly is suspended at Us midpoint by a wire and rotates in a horizontal plane, making 3.0 resolutions per minute, find the angular momentum and kinetic energy of the assembly.

Explanation / Answer

no no no u shouldnt take 2/5mr^2,

for this problem consider sphere as point mass and than from centre calucate its MOI

GENERAL MOI FORMULA = m*r^2

where r=s/2

so for our spheres MOI AROUN ROD AXIS IS = m*(s/2)^2

two sphers at both end so I spheres = 2*m*(s/2)^2 = 16 kg m^2

and for our rod its = 4 kg m^2

total = 16 kg m^2 + 4 kg m^2 = 20 kg m^2

always remeber MOI IS always about axis of rotation

and I = 2/5mr^2 is the formula when axis of rotation passing through sphere which is not in our case .

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