-
Looking for HSC notes and resources? Check out our Notes & Resources page
Volumes and Mechanics questions (1 Viewer)
- Thread starter pikachu975
- Start date
pikachu975
Premium Member
I tried integrating so like acceleration is $a = -\frac{V_{0}^{2}}{2l}$. The force is thus $ma = -\frac{mV_{0}^{2}}{2l}$, so the answer is (C).$)
a = -F/m
and vdv/dx = -F/m
dv/dx = -Fv/m
dx = -mdv/Fv
[x](0 to L) = -m/F [v^2 / 2](Vo to 0)
L = m/F (Vo^2 / 2)
F = mVo^2 / 2L
But idk what I did wrong here
pikachu975
Premium Member
For the volumes I tried I keep getting 16 instead of 8 because I did:. Note that the last two integrals can be immediately ruled out as they are $0$ (integrating an odd function over a symmetric-about-0 interval). So if you had no idea and had to guess, you could narrow it down to between (A) and (B). Anyway, one way to see that the answer is (B) is to show that this is the integral that results if we try to find the volume using cylindrical shells.$)
dV = 2pirh dy
= 2pi * 2(1-y) * (2x)
= 8pi (1-y)(2sqrt(1-y^2))
= 16pi (1-y)(sqrt(1-y^2))
Idk where I went wrong
It's not 2(1-y) but 1-y
Basically the first line – it should be a = F/m. (Since F = ma.)
pikachu975
Premium Member
Why is the radius only 1-y? Wouldn't it be 2(1-y) since 1-y only covers half the ellipse\times 4\sqrt{1 - y^{2}} = 8\pi(1-y)\sqrt{1 - y^{2}}$.$)
It doesn't cover half the ellipse. It's basically the shift of the curve. That's why it's (1-y) as the radius
pikachu975
Premium Member
Thanks heaps!
If it wasn't a multiple choice, are you allowed to use physics formulae?
Generally speaking, no.
pikachu975
Premium Member
You'd use a = F/m and a = vdv/dx
why isn't it a=-F/m. The force is opposing the motion
We always have F = ma where F is the net force. The fact that the force is opposing the motion is reflected by the fact that F and a are negative.
Last edited: