Calculating Mass Moment of Inertia for a Rotating Rod | Homework Help

[reckk]

hi.i've been asked to post this question here for a concrete answer..

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

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

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?

 

[kataya]

For the first part you need to use the parallel axis theorem. It states the following:

$$I_{p}=I_{cm} + mk^{2}$$

Where $$I_{p}$$ is the mass moment of inertia about any point, $$I_{cm}$$ is the mass moment of inertia about the object's center of mass, and k is the distance between the object's center of mass and the axis of revolution.

Clean up some of the units before you post.

total mass of the rod = 4m
m = 0.4kg; b = 0.4m; a = 1.0m, n = 800 U/min
rod is thin

I'm going to assume that you meant that the mass of the rod is 4kg and the speed of rotation is 800rpm? I am not really sure about that last unit.

Anyways, there will be no moment on the rod (and thus torque on the bar) unless it is accelerating. If it is simply rotating at a constant velocity then the only thing you need to take into consideration is the weight of the rod. Maybe I could help you more if you explained your reasoning verbally.

 

[reckk]

mass of the rod is 4m.. so the total mass would be 0.4x4 = 1.6 kg..
yup..revolution per minute is 800 rpm.. sry about dat..
and the rod is homogeneous..

for the 2nd question..
it's only the rod which is parallel to the z-axis is to be considered rite?
so to calculate the moment about x axis, i should consider the center of mass of the rod parallel to z-axis..

is it correct?