Recent content by Stickboy10

why are they all squared when the original equation was v = r\omega? where does the squaring come from? oh, omega is sqaured in the conservation equation I set up so you just moved it into the other equation to show an easy variable switch.

Ah, ok I think I see what you are saying and I see how the relationship brings in the r which something I have to have in my final equation. So I have: MgH = \frac{1}{2}I\omega2 + \frac{1}{2}Mv2 and \omega= v/r so MgH = \frac{1}{2}I\frac{v}{r}2 + \frac{1}{2}Mv2 I don't know if this...

Oh duh! I don't why I didnt see that! Thank you AlexChandler.

I see the relationship in the equation between v and \omega but from there I don't see how I am relating the r with the work that I have done so far.

Okay so from this I find that \bar{L} = \bar{r} x m\bar{v} and \bar{\tau} = \bar{r} x m\bar{a} The difference between the two is the acceleration from the velocity. Acceleration is the derivative of velocity. So it looks like I find the linear momentum first and then since I have that I...

U=MgH \leftarrow since I don't have numbers I can't calculate that. but I know Ui + Ki = Uf + Kf (assuming no E loss) \leftarrow where Ki is zero and Uf is zero. I can split kf into Krot and Ktrans so U + 0 = 0 + Krot + Ktrans Ktrans = \frac{1}{2}Mv2\leftarrow inserting my vtrans in gives...

Hi guys, I am hoping you can either point me in the right direction here or show me how to do this a bit. the problem is as follows: "a particle with weight 86.8 N is positioned at r= (8.1t)i - (7.2t-9.4t2)j. t is in seconds. Find an expression for angular momentum, L and torque, T which act...

Hello there I am really struggling with this and another problem and I was hoping you guys could help me out. The problem is as follows: "the cylinder shown has a radially-dependent density with mass M and radius R. The cylinder starts from rest and rolls without slipping down the incline...