Recent content by NTesla

I don't understand why it doesn't make sense. In undergraduate level physics books, that's how angular velocity and angular acceleration has been introduced. There's always an axis about which these terms are defined. Kindly let me know your opinion on this.

I understand your arguments and the calculation. However, In view of that wikipedia statement, and your arguments and calculation, I'm trying to find out a common ground or a finality regarding the exact process of calculation of Moment of Inertia, and it's definition too. But it seems so...

I must admit, I can understand each word individually, but I couldn't understand what you wanted to say there, at all. Kindly explain in simpler, preferably much simpler terms, what you wanted to write.

Kindly see post#57 and 58. The conclusion that I'm gathering from those 2 posts is this: The net angular accleration about point C will be: ##\vec{\alpha }_{net} = \vec{\alpha}_{cm} + \vec{\alpha}_{C}##. And since we have the relation between ##\vec{\alpha}_{cm}## and ##\vec{\alpha}_{C}##...

I would like your response on this question.

I've figured out where the sign convention went wrong, and have arrived at the correct answer. @haruspex, @kuruman : Appreciate your kind help very much. However, now I'm working on another method of solving the same problem of finding the time period of oscillation of the ball, by using...

I've figured out where the sign convention went wrong, and have arrived at the correct answer. @haruspex, @kuruman : Appreciate your kind help very much. However, I'm working on another method of solving the same problem of finding the time period of oscillation of the ball, by using Energy...

yes, and it is already negative, before we proceed on to do factorization. Kindly see last line in post#59. So, eventually ##\alpha_{C}## is coming out positive, after the factorization, which it shouldn't.

After factorizing, in the denominator I'm getting: ##(r - R)##, where ##(R - r)## should come. I've double checked my calculations/equations, but can't find why the sign is reversed.

I don't understand how does that reduce to this equation. Even if we neglect the small r term in the numerator, we can't neglect the term 12rR in the denominator.

Here's the calculation taking the angular momentum about point C. ## L_{aboutC} = -mw_{C}(R-r)^2 + \frac{2}{5}mr^2w_{cm} ## and using the equation: ##w_{cm} = \frac{R-r}{r}w_{c}##, we get: ## L_{aboutC} = \left ( -R^2 - \frac{7}{5}r^2 + \frac{12}{5}rR\right )mw_{c}## Differentiation both sides...

Yes, i agree that that is what you were saying. I'm not sure why you thought I was disagreeing with you on this point. In the statement: "He is saying that parallel axis theorem can't be applied as you and I have done so far." "as" is the operative word. Meaning that you are saying that PAT...

Not really, because I already do understand that. In my calculation that I have attached as pdf in post#24, I have already written that ##I_{about C} = \frac{2}{5}mr^2+m(R-r)^2.## and then have proceeded to calculate the time period. Thereafter, in my post#43, I again mentioned that: In that...

Agreed. I don't agree. There is a definite non-zero moment of Inertia about C, and finding it could possibly lead to increase in knowledge about the whole thing.

I'm not sure how that would help in finding out the moment of Inertia of the ball about the axis passing through point C, in the present question(in case of Pic 2 of post#43).