Recent content by Eggue

I honestly have no idea where I'm going wrong i've checked my differentiation using a derivative calculator but the answer is the same. The answer from the book is \frac{L}{2} + \frac{h^2}{L} Working:

Ah i understand now. Thank you

No what i meant was the sign on a_x changes the value of t and i do not know why the sign should be positive even though friction is pointing in the -x direction.

Oh that was a mistake that i corrected later, but that shouldn't affect anything right?

I'm not sure as to why my working is incorrect. When the sign on a_x is positive, i get t = \frac{R\omega_0}{3\mu_kg} which would give the correct value for distance if plugged into the kinematic equation. However, I'm not sure why a_x would be positive though since the friction force is...

I got the correct answer for the first part but I'm not sure why the answer for (b) is the same for (a). Wouldn't the rings falling off mean that I_f = \frac{1}{12}M_L L^2 only where I_F, M_L, L are the final moment of inertia, mass of the rod and length of the rod as opposed to I_f =...

I did a problem before that asked to find the moment of inertia of the 2 rod system so i sort of had tunnel vision

Ohhh. Thank you so much for the help I've been trying to figure out where I've gone wrong for 2 days

Haha sorry about the image quality. You're right, once i correct the error i get the correct answer. One more question though, Why does the center of mass of the 2 rods lie on the line connecting the center of mass of each rod? Shouldn't it lie halfway between both rods which would give a...

I did the working but my answer isn't correct Working: Where I_p is the moment of inertia about the center of the square

So will the distance to the center of mass be \frac{\sqrt{2}a}{4} ?

Aren't they the same though?

I tried to find the moment of inertia of 2 rods connected at the corners by adding up their moments of inertia: \frac{1}{3}(\frac{M}{4})a^2 + \frac{1}{3}(\frac{M}{4})a^2 = \frac{1}{6}Ma^2 I then tried to solve for the moment of inertia at the center of mass of the 2 rods using the parallel...