Recent content by shishykish

Hi! So I should be thinking conceptually in terms of calculus. Taking an elementary shape then summing over the whole object is one of the most basic approaches.

Ha! Yes it is. I really could learn how to copy and paste properly!

radius of gyration for a uniform rod is $$ k= \sqrt{ \frac{a^3}{3} }$$from what I remember of deriving it from my notes. So we are actually taking this shape as an elementary strip between the limits of 2 to 4. The total moment of area will be the $$ I_{area} = area \times (radius of gyration)^2 $$

Hi, I'm new here as a physics teacher. I've been teaching since 2007. A lot of students have a lot of trouble with relating the physical reality to terms in an equation. I find analogies and models very effective in helping students. For example, the friction formula is a good one. As you...

Area of strip = $$/delta x /times y$$ Moment of area for strip $$= Area /times (radius of gyration)^2 $$ $$ = /delta x /times y /times (/frac{y^2}{3})^2 $$ Why is the radius of gyration squared term $$y^2$ over 3? $$