Gauranga
- 14
- 0
Guys i need a little help in finding the moment of inertia of a part of a machine which looks like the part of a torus as the arc of a circle.Can somebody help me out with it please?
Moment of Inertia for a machine
- Thread starter Gauranga
- Start date
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
22 replies · 12K views
Discussion
Admin
- 69,782
- 25,629
Gauranga said:
Welcome to the PF.
What is the context of the question? Where is this part used, and why do you need its moment of inertia?
What is your math and engineering background? Are you familiar with the basic equations for calculating the moment of inertia?
http://en.wikipedia.org/wiki/Moment_of_inertia
.
Gauranga
- 14
- 0
I am an engineer doing and i know the basic equation but the problem is with the part.Basically it is an arm of an instrument which moves and i have to calculate the moment of inertia of it . But the form of it is a little complex. It is a hollow section of solid steel which is similar to a section of a torus.I tried to find formulas of moment of inertia for different structure but the one i need like a 3d hollow arc could not be found.So it will be great if u can help me in that. Ok may be it could be described if u cut the one third of a torus and want to find the moment of inertia of it.Where the arc length is given.the torus diameter is given.The radius of the torus from the center is given to find out the moment of inertia.May be it can be related to some formula?I just can't get it...
W R-P
- 25
- 0
gsal
- 1,065
- 54
if you do not find a formula, you may need to go back to the basics and calculate the moment of inertia yourself...
Last edited by a moderator:
Gauranga
- 14
- 0
I think i will calculate the moment of inertia by the formula of a hollow torus and then divide it to 8 as the angle between the two radius connecting the arc is 45 degree.but for this i have to take a assumption that this body is a homgeneous torus because there are some slots and additional structure for bolts,screws and all.
Gauranga
- 14
- 0
Hey.I am learning the solid edge and i didn't know that i can find out the moment of inertia in that.However thanks a lot.
Gauranga
- 14
- 0
Hey guys as i found out the moment of inertia and the center of mass of the machine, now the problem is to find our the torque.So i have 300rpm and I = 180000 gm*mm^2. but i can't relate between these two quantity to find out the torque.can somebody provide me some relationship.Is it so that torque = I.omega.2pi/60. As i read some information about it but i am not sure if i got it right. where omega i mean by the angular velocity.
xandro101
- 22
- 0
Torque is I* angular acceleration.
Gauranga
- 14
- 0
Hehehhe thanks a lot for the information.I was looking for something more than that.So that i can find the torque from above information but i understood that i need the acceleration time which i have to get.Thanks by the way.
xandro101
- 22
- 0
Yes, u have to get time of acceleration to that speed. Also, is the acceleration uniform?
Gauranga
- 14
- 0
So i have the uniform accelaration.My rpm is 3.6,acceleration time is half second and my moment of inertia is 71502884 gm cm^2 and in case i have the mass also 75.8 kg.Can somebody help me to use these values to find the torque.
gsal
- 1,065
- 54
Yes, somebody could...but while the original reason for this thread might have been understandable and you might be all into it by now...don't give up on the net...can you just google something by yourself? like your last question? come on...
Gauranga
- 14
- 0
i think i have found it out by founding angular accn 0.75 rad/sec^2 and by the general formula torque = moment of inertia * angular accelaration.The result is 5.39 kgm^2 rad/sec^2 where i think the rad is not the mater.So the unit seems good.But i was not sure.can someone say that the method was right?
xandro101
- 22
- 0
How is your angular acceleration 0.75rad/s^2 ? You have an rpm of 3.6 being reached in half second. Going by the formula for angular acceleration which is "rpm divided by time taken" it should be 3.6/0.5 = 7.8rad/s^2. Let me know how u got 0.75rad/s^2.
Gauranga
- 14
- 0
First of all i changed the rpm 3.6n into rad by the formula omega=2*pi*3.6/60. And then i divided the time with it to get the angular acceleration.
Gauranga
- 14
- 0
gsal said:
I am all into it and it is the part of my last question which is proceeding further and further as i have lots to do about it.However thank.
xandro101
- 22
- 0
Yea. I understand now. Thanks.
Gauranga
- 14
- 0
Guys can you please just check the method i applied for torque calculation.
First of all i found the moment of inertia.Then i have the value of a stroke, from where i calculated the rpm.
Then by the formula 2.pi.rpm/60 ,i got the angular acceleration in rad/sec.So i devided the acceleration time by it and got the angular acceleration in rad/sec^2 .
I multiplied that with moment of inertia which was in kg m^2 and got a value in kg.m^2rad/ sec^2.
Where i think rad has no play.so the resulting unit will be Nm which is the right unit for torque.
Can somebody just have a look on the method and say me if i am right?
First of all i found the moment of inertia.Then i have the value of a stroke, from where i calculated the rpm.
Then by the formula 2.pi.rpm/60 ,i got the angular acceleration in rad/sec.So i devided the acceleration time by it and got the angular acceleration in rad/sec^2 .
I multiplied that with moment of inertia which was in kg m^2 and got a value in kg.m^2rad/ sec^2.
Where i think rad has no play.so the resulting unit will be Nm which is the right unit for torque.
Can somebody just have a look on the method and say me if i am right?
- 12,179
- 186
Gauranga said:
Gauranga said:
Looks good.Gauranga said:
And you're correct about the units, as "rad" is considered unitless.
Gauranga
- 14
- 0
Thanks...