Recent content by royguitarboy

Ok. I think I might have it. So for C, I need to add what I got in part B to the cross product of v and p, and subtract it, cause there should be two answers. Is that right? Well, I'm kind of confused about how to exactly do the cross product, we didn't really go over that.Edit: Never mind...

I'm confused as to which r, I should use for each part. In the the r X p part, the r is .05m right? So the total momentum is: I(omega) + r X p so I got 2.34e-5 + .05m X (.011)(\omega*(.019/2) ?

ok I got you now. Now part B, I need to add part A to \vec{L} = \vec{r} \times \vec{p} ?

Ok, now I'm really confused about this. For part A, I need to use \vec{L} = \vec{r} \times \vec{p} where p is mass times velocity?

L = I \omega just gives me the angular momentum about the center of mass though right? How do I account for a distance 10cm from that point?

Homework Statement A 11 g coin of diameter 1.9 cm is spinning at 15 rev/s about a vertical diameter at a fixed point on a tabletop. A coin is a solid cylinder of length L and radius R, where L is negligible compared to R. Its moment of inertia is 1/4MR^2. (A)What is its angular momentum...

Ok, I have the first part now. For the second, I get that the T=m1(g+a) and T=m2(g-a), and then that a=(m1g-m2g)/(m1+m2). Is that even close for (b) If 12 kg is gently added to the top of m1, find the angular acceleration of the wheels ?

I'm sorry I'm having a hard time understanding this; The tensions must cancel each other since there is no acceleration right? Maybe if it's explained in another way I might get it.

I understand that the tensions equal mass times gravity. I need to figure out what the second mass is, and I'm pretty sure I do this by figuring out how the tensions are related.

So how do I go about figuring out the tensions?

Wouldn't the tensions equaling mg mean that m2 would need to equal m1?

I know that the overall net Force is 0 because there is no acceleration. All I get when I apply Newton's Laws is F=T-mg. This would mean that the Tension equals mg, which I don't think is true. I think I need to figure out the relationship between the two tensions, but I'm not sure how.

Homework Statement Two objects are attached to ropes that are attached to wheels on a common axle as shown below. The two wheels are glued together so that they form a single object. The total moment of inertia of the object is 41 kg·m2. The radii of the wheels are R1 = 1 m and R2 = 0.4 m. (a)...

Sorry I'm a little confused. I'm not sure how to get the velocity at the lowest point other than it being sin(0)*1.6 which would be 0. And doesn't " sin(5.73)*1.6*.5*9.8 + m2v2f^2" go into the momentum equation? If it's easier to talk on aim, my sn is: Royguitarboy246 Edit: Wouldn't that...

Ki(objec1t)+Ki(object2)+Ui(object1)+Ui(object2)=Kf(object1)+Kf(object2)+Uf(object1)+Uf(object2) K= kinetic energy------U= potential energy-------i=initial------f=final K=1/2mv^2-----------U=mgh 0+0+0+cos(53)*1.6*.5*9.8=sin(5.73)(1.6)+(.5)(cos(5.73)*1.6)+m2vf^2 Change in kinetic energy = 0 so...