Recent content by Raghavendar Balaji

Oh thank you! I think that might work! Thanks for the suggestion!

Exactly my point. It's highly risky but it can be tried though it might not be possible in all directions. So I'm out of ideas.

Hi, thanks for the suggestion. In case of an engine or a gearbox, there are lots of components so it's totally complex but balancing can be tried but, do you have any other method or suggestion to solve this problem?

Homework Statement A complex or irregular body with some mass, m. How to determine the 3 co-ordinates of centre of gravity? Homework Equations Moment method with ∑M = 0. The Attempt at a Solution Say I have a gearbox and I need to find the COG. I can mount the gearbox at two longitudinal...

Oh okay! Is it possible to project the disk onto a horizontal plane which would result in an ellipse and then use the parallel axis theorem to it? Or if not, how can this be dealt with?

Hi! I did it from 0 to 2.pi and got the right value! Thank you very much for the invaluable help! I'll never forget this anytime, now! :) Just a small question, we found out the moment of inertia about an axis through the COG. So for the actual problem, since it is offset from the center, can I...

Yes, then it would be ρ∫((r.cos(Φ).cos(α))^2 + (r.sin(Φ))^2).r.dr.dΦ, with Φ ranging from 0 to π/2 and r from 0 to R?

Oh sorry! The distance would be √((ysinβ)^2 + z^2). Is that it ?

Ohh yes! So it would be the sum of squared distance of y and z co-ordinate. Am I right?

I guess only the square of y coordinate line, since z remains constant and x = 0. Am I right?

Oh yes, I'm sorry, the y coordinate line comes closer and not the axis whereas the z coordinate line remains constant throughout the rotation. :)

When the vector is kept normal to the lines i.e., perpendicular to the surface of the paper, the y-axis comes closer to the pen when rotated about the z axis. Is it right?

When the inclination is β, then for both the co-ordinates, would it be multiplied by sinβ? Sorry if I am wrong :(

Yes I agree, a point on the circumference of the disk would have the mentioned coordinates which changes with the location of the point, but the hinged angle would remain constant. Right?

:D sure will do! The polar co-ordinates with angle to the vertical would be y = r.cosΦ and z = r.sinΦ and if the angle is w.r.t horizontal, it would be the reverse. Is it right?