moment arm = h-y
1/2 * integration of (density *g * L)(y)dy * (h-y)
so integration of density * g * l * y dy * h - integration of density * g * l * y dy * -y
so 1/2 density * g * l * h^3 - 1/3 density * g * l * y * h^3
= 1/6 density g * L * h^3!
Integration of torque = integration of dF * moment arms
so 1/2 * integration of (density *g * L)(y)dy * y??
thats the best I got, I hope thats (finally) right.
end up with 1/6 density * g * L *h^3
I worked with someone from my class on this problem and they said they ended up with this:
1/2L*density*g*h^3 - 1/3L*density*h^3 = 1/6L*density*g*h^3
They're pretty positive its right and I see where they get what is subtracted but how did they find 1/2L*density*g*h^3??
I thought I already posted the torque from each strip? 1/3(density*g*L)y^3?? (at least thats what I guess) If I'm finding torque with respect to the bottom, then that is the highest positive y value. I guess I'm confused as to where you are taking me here?? Should I not post that integration...
Homework Statement
Realize I'm asking a lot of question today, but I'm working on a huge physics project and I'm not asking for answers, just some guidance where I'm stuck
A certain yo-yo can be modeled as a uniform cylindrical disk with mass M and radius R and a lightweight hub of radius...
Is this a legit way to go about it?
Initial velocity gives initial kinetic energy KE = 1/2 * Mmars * V^2, where Mmars is the mass of mars.
Some of that energy goes into melting rock. However, depending on the impact, maybe only a small amount of rock was vaporized and the majority of rock...
Ohh hold on. So y would be zero at the surface of the water (y is the thickness) and it gets larger as you go down. Maybe this has to do with h? so maybe I'd have to convert this to h again like in the last problem?
so 1/3(density*g*L)h^3??
Fantastic. That's a really good way to look at it. I think I'll set r_e pretty far from the center as I believe that the impact occured near the edge of the earth.
So to find the velocity, I just need to find the angular momentum of the earth moon system afterwards. But from what I...
Oh so you're saying that h and 0 are the limits of the integration so when its integrated, I should get '= 1/2 (density *g * L)(h^2-0)??? I think that makes sense
In regards to where y is measured from, I guess y is zero at the bottom? Hmm I don't really know where y is measured from
Alright so if I used y instead of h, that is okay?
so For part a)
break dam into little strips, each strip has a dif force...
dF = PdA = density(g)(L)(y)dy <-- integrate this?
= 1/2 (density *g * L)(y^2)???
and that seems about right? I think I can forget about atmospheric pressure...
Homework Statement
One likely theory for the origin of the Moon is that it was formed by the impact of a Mars-sized object on the young Earth. How fast would the object have to be going when it hit the Earth to vaporize enough rock to make the Moon?
Homework Equations
The Attempt...
Homework Statement
Consider a simple model of a dam. A reservoir of water behind the dam is filled to a height h. Assume that the width of the dam is L. As the reservoir behind a dam is filled with water, the pressure that the water exerts on the dam increases. Eventually, the force on the...
In a similar vein, I was trying to figure out what would happen to the length of our days if the moons distance was 1.5 times its current value from earth. I found an equation online that said that P = A^1.5 where P is the orbital period and A is the average distance. I don't understand how they...