0
$\begingroup$

Hi so here's the problem I'm working on:

Consider a 2 meter-long rod with linear mass density $\lambda(x)=x$ kg/m, where $x=0$ corresponds to one end of the rod.

(a) Determine the centre of mass of the rod.

(b) About which end ($x=0$ or $x=2$) is the rod easier to rotate? Justify your answer with a detailed calculation.

I'm not currently in a physics class that has dealt with problems involving stuff with centres of mass and moments of inertia, so excuse my terminology if it's a bit off (sorry in advance). So I've already found the answer for (a), in which the centre of mass is $\frac{4}{3}$. My only problem is that I'm not sure if I am approaching part (b) correctly. If I'm correct, I calculate the moment of inertia as $I = \int\limits_{0}^{2}x^2\lambda(x)dx$, right? If I do this, my moment of inertia about the $x=0$ end gets me $I = \int\limits_{0}^{2}x^3dx$ and $I=4$. Now what's confusing me is how would I calculate the moment of inertia at the $x=2$ end? If I do the same thing I did for calculating the moment of inertia at the $x=0$ end, I get $-4$ as my moment of inertia which would mean at both ends, rotating takes the same amount of "effort". However, wouldn't this not make sense since the centre of mass of the rod is $\frac{4}{3}$ and not at the centre of the rod? Or is the centre of mass of the rod completely irrelevant to solving this part of the problem?

$\endgroup$
0
$\begingroup$

you have to calculate the moment of inertia about the centre of mass first, which is, let us call this $I_{CM}$, using the parallel axis theorem you get

$I=I_\text{CM}+md^2$ where $d$ is the distance to the centre of mass from where you are calculating the moment of inertia, and $m$ is the whole mass

so from the point $x=0$:

$I_0=I_\text{CM}+m(4/3)^2$

and from point $x=2$:

$I_2=I_\text{CM}+m(2-4/3)^2$

$I_2=I_\text{CM}+m(2/3)^2$

you can see that $I_2$ is less than $I_0$ therefore rotating about the point $x=2$ is easier.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.