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reckk
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hi.i've been asked to post this question here for a concrete answer..
the above question has been solved which the answer is 0.1706
and here's the another question that i need a help
and here's my another approach
reckk said:Homework Statement
http://img232.imageshack.us/img232/4976/physicswe5.jpg
total mass of the rod = 4m
m = 0.4kg; b = 0.4m; a = 1.0m, n = 800 U/min
rod is thin
i need to find the mass moment of inertia for the rotating rod relative to z-axis..
Homework Equations
(1/12)ML²
The Attempt at a Solution
for the rod which lies on y axis, i could calculate the mass moment of inertia by using the above equation which will lead me to the following answer
(1/12)*2*0.40*0.8² = 0.0427
but what should i do with the other parts of the rod which are parallel with the z-axis?..
or could i apply the following equation?:
(1/24)*M*L²*sin (2φ)
where φ = 45
thx in advance..
the above question has been solved which the answer is 0.1706
and here's the another question that i need a help
reckk said:here's the last question..
i need to find the bearing reaction on A and B
value given : n = 800/min; m = 0.4kg; a = 1.0m; b = 0.4m
here's my approach to solve the question:
Moment about x-axis:
Mx = 2 Jyz*ω²
= 2ω² ∫ b*(b/2) dm
= 2ω² ∫ b²/2 dm
= ω² * (b²/2) * m
ω = 2(pi)n/60 = 83.776 (1/s)
Mx = 83.776² * 0.4²/2 * 0.4
= 449.18 Nm
FA = FB =
Mx/2a
= 224.59 N
i wonder if i have done the right approach.. i took b/2 as its center of mass.. so i came up with following equation
∫ b*(b/2) dm
and since there's two parts which is parallel to z-axis.. i time the mass moment of inertia with 2..
is this the way to answer the question?.. I'm kind of confused with another method to calculate moment of inertia where the rod has an angle φ to the y-axis..
and here's my another approach
reckk said:i came up with another approach
J = Jz1 + Jz2
= 2mr²
= 2*0.4*0.4²
= 0.128
Mx = Jω²
= 0.128 * (2(pi)n/60)²
= 898.35 N
FA = FB = Mx/2a
= 449.175 N
so..which one is the right approach? or is there any another approach?
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