Recent content by Mercy

Hi all, I was hoping some of you had experience with this, I'm taking a gap year between undergrad and grad and was hoping to keep myself occupied during the Fall and Spring seasons, although I'm looking to keep doing research rather than working for starbucks. What options are there? I've...

Ah I see, in math that is known as the power series of sin(x). When you expand the series it becomes: $$ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + ... $$ and that goes on to infinity. So when you take y = x - sin(x) it cancels out that first x term in the series and...

haha fair enough, do you remember what the answer you were satisfied with was?

Is it because they are linearly independent? Like how if you have y = sin(x) + cos(x) you can't isolate x? In my head I just expand out the power series for sine and then you end up with an infinite degree polynomial which can't be reduced, not sure if that's the answer though

Did you also learn about the cross product? Torque is defined as the radius vector crossed with the force vector. You've learned about translational motion; F = ma (a constant force causes an object to accelerate translationally), analogously, a constant torque causes an object to accelerate...

I would assume so, you would just be taking another set of time derivatives, although I'm not 100% sure on that.

The $$ (\dot{\omega} \times r) $$ term comes from the product rule from the time derivative of the last term in the first line.

No, what I wrote is what is in my textbook (Thorton/Marion pg 389 - 392). Fixed frame quantities expand this way: $$ \big( \frac{dr}{dt} \big)_{fixed} = \big( \frac{dr}{dt} \big)_{rot}+ (\omega \times r) $$ So from the second line: $$ \big( \frac{dv_{rot}}{dt} \big)_{fixed} =a_r+ (\omega...

haha thank you it was bugging me that I couldn't figure it out :)

$$ v_{fix} = \frac{dr}{dt}_{rot} + (\omega x r) $$ $$ \big( \frac{dv}{dt} \big)_{fix} = \big( \frac{dv_{rot}}{dt} \big)_{fix} + (\dot{\omega} x r) + \omega x \big( \frac{dr}{dt} \big)_{fix} $$ $$a_f = a_r + (\omega x v_r) + (\dot{\omega} x r) + (\omega x (\omega x r) ) + (\omega x v_r) $$ $$...

Then did you forget to include the last term in the second to last line? It looks like it disappeared

Your moment of inertia tensor is only true for where you center your origin. In most cases centering your coordinate system at the center of mass and calculating the tensor from there is complicated. The usual method for these problems is first calculating your inertia tensor in a "convenient"...

In the last line where you cancel out $$ \frac {d}{dt} (\omega x r) $$ Use the product rule on this term to obtain $$ (\dot{\omega} x r) + (\omega x \dot{r}) $$ the former term is the one that you throw away and the latter, combined with the last term in your second to last line, form...

It's hard to know what is and what isn't a meaningful question when the answer could be beyond the scope of what I've learned. I could analogously ask "what is the relationship between the heat of a system and the system's specific heat capacity". One is a constant and one isn't a state...

Nasu, the fact that one is a constant and one is a variable doesn't mean they aren't related in some way. In the case of c and v, yes they share units but there are also physical relationships between the two such as 'an object's speed can never reach c', and of course the relationship between...