Where do I place the bearings for a stable inclined rotating drum?

AI Thread Summary
The discussion centers on determining the optimal placement of bearings for a rotary drum dryer, specifically focusing on stability given the drum's dimensions and incline. The drum measures 17 m in length and 3.24 m in diameter, inclined at 2° and rotating at 3.24 rpm, with a weight of 280 kN. Participants emphasize the importance of accurately finding the center of mass, which remains at the midpoint when inclined, and suggest that the moment calculations should consider the uniformity of the drum. There is some confusion regarding the specific type of bearings needed and the relevance of the slight incline. Overall, understanding the center of mass and its implications for bearing placement is crucial for the drum's stability.
Sook
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Homework Statement


I am basically designing a rotary drum dryer and need to know the positions of the bearings for the dryer to be stable.
The drum is of length 17 m with diameter 3.24 m, inclined at 2° rotating at 3.24 rpm. I need to find the distance X and OY. (Given that Y passes through the center of mass) Weight of the drum =280 kN

http://https://scontent-a-kul.xx.fbcdn.net/hphotos-ash3/t31.0-8/p417x417/10014962_10152301201595569_2033873904_o.jpg

Homework Equations



Moment= F x D = WD ;
Moment CW = Moment ACW


The Attempt at a Solution


I am trying to solve this using moments. But I am stuck with finding the centre of mass of the inclined cylinder.

According to me (and after reading alignments of rotary drum, they did not take into consideration the inclination), X can be found by : (17/2)/2 =4.25m (Is that possible?)
 
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Not sure I will be around over the next few days to help but I suggest you post a diagram providing more detail such as roughly where the bearings are.

Are these the bearings that the drum spins in OR the bearings that the drum pivots about when changing inclination to tip the contents out or ??

2 degrees is quite a small angle.. not far off horizontal..so why do you think it needs to be taken into account?

I am stuck with finding the centre of mass of the inclined cylinder.

Finding the centre of mass of an inclined cylinder is quite trivial. So I think you might be asking the wrong question?

Lets say that the drum is uniform and the centre of mass is mid way between the ends when viewed horizontally (eg 17/2 = 8.5m from one end). Then when inclined it will still be mid way between the ends when inclined. See diagram.

A = 0.5*17*Cos(0) = 8.5m
B = 0.5*17*Cos(2) = 8.4948m

Similarly if the centre of mass is say 1/3rd of the way from one end.
 

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PS. Just noticed you tried to post a diagram but I can't see it or open it.
 
This being the homework forum, PF rules stipulate that only assistance to understanding should be given, not a complete solution to any problem posted.

Sook's pic:

https://scontent-a-kul.xx.fbcdn.net/hphotos-ash3/t31.0-8/p417x417/10014962_10152301201595569_2033873904_o.jpg
 
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