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

ID: 1311259 • Letter: O

Question

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

my problem while solving this was that before there was exact same question like this one

but the solid cylinder were known as point masses. so i didnt have to care about sold cylinder =1/2mr^2

however, the first step of this is get sum of Inertia and since i know omega w , convert that into rad/s and put in to equation L=iw so easy..but

i was confused when i thought, the i sphere= 1/2mr^2 so 1/2 (2kg)(2m)^2 =4.. i put 2 because from axis of rotation to sold cyliner the radius would be 2 not 4... !!! and since there are 2 cylinder i just have to add one more or multiply 2 so 1/2 (2kg)(2m)^2+ 1/2 (2kg)(2m)^2 = give me 8kg m^2 not 16kgm^2

in the book it seems like they used 1/2 (2kg)(4m)^2 but wth? arent i suppose to use 2m? because of axis of rotation is in the middle.!!

so total would be Inertia of rod+ 2balls of solid cylinder....

help confused with radius...

because last time it was correct, it was exact same problem. but cylinder was known as point mass.. and even my professor said radius from axis of rotation to the ball would be 1/2 of orginal radius.. !

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

you have got your formula wrong

The moment of inertia of a point mass = mr2 (where r is the distance from the axis to mass) and not 1/2 mr2 ?and in this case the spheres at the end of the rod are considered as point masses.. so there is nothing wrong in the explaination given.

in the book they used i = m (length of cylinder / 2)2  which is effectively mr2

Infact moment of inertia of sphere (hollow) = 2/3 mr2 (r radius of sphere)

sphere (solid) = 2/5 mr2 (r- radius) and never 1/2 mr2