# Thread: Volumes and Mechanics questions

1. ## Volumes and Mechanics questions

http://prntscr.com/ffbno6 - mechanics

https://prnt.sc/ffa8xe - volumes

Ty for any help!!

2. ## Re: Volumes and Mechanics questions

Originally Posted by pikachu975
http://prntscr.com/ffbno6 - mechanics

https://prnt.sc/ffa8xe - volumes

Ty for any help!!
$\noindent For the mechanics one, using v^{2} = u^{2} + 2as with v=0, u = V_{0}, s = l, we can solve for a to find that the required (constant) acceleration is a = -\frac{V_{0}^{2}}{2l}. The force is thus ma = -\frac{mV_{0}^{2}}{2l}, so the answer is (C).$

3. ## Re: Volumes and Mechanics questions

Originally Posted by pikachu975
http://prntscr.com/ffbno6 - mechanics

https://prnt.sc/ffa8xe - volumes

Ty for any help!!
$\noindent The volumes one is (B). Note that the last two integrals can be immediately ruled out as they are 0 (integrating an odd function over a symmetric-about-0 interval). So if you had no idea and had to guess, you could narrow it down to between (A) and (B). Anyway, one way to see that the answer is (B) is to show that this is the integral that results if we try to find the volume using cylindrical shells.$

4. ## Re: Volumes and Mechanics questions

Originally Posted by InteGrand
$\noindent For the mechanics one, using v^{2} = u^{2} + 2as with v=0, u = V_{0}, s = l, we can solve for a to find that the required (constant) acceleration is a = -\frac{V_{0}^{2}}{2l}. The force is thus ma = -\frac{mV_{0}^{2}}{2l}, so the answer is (C).$
I tried integrating so like

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

5. ## Re: Volumes and Mechanics questions

Originally Posted by InteGrand
$\noindent The volumes one is (B). Note that the last two integrals can be immediately ruled out as they are 0 (integrating an odd function over a symmetric-about-0 interval). So if you had no idea and had to guess, you could narrow it down to between (A) and (B). Anyway, one way to see that the answer is (B) is to show that this is the integral that results if we try to find the volume using cylindrical shells.$
For the volumes I tried I keep getting 16 instead of 8 because I did:
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

6. ## Re: Volumes and Mechanics questions

Originally Posted by pikachu975
For the volumes I tried I keep getting 16 instead of 8 because I did:
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

7. ## Re: Volumes and Mechanics questions

Originally Posted by pikachu975
For the volumes I tried I keep getting 16 instead of 8 because I did:
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
$\noindent The r should just be 1-y and the h should be 4\sqrt{1-y^{2}}. So 2\pi r h = 2\pi (1-y)\times 4\sqrt{1 - y^{2}} = 8\pi(1-y)\sqrt{1 - y^{2}}.$

8. ## Re: Volumes and Mechanics questions

Originally Posted by pikachu975
I tried integrating so like

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
Basically the first line – it should be a = F/m. (Since F = ma.)

9. ## Re: Volumes and Mechanics questions

Originally Posted by InteGrand
$\noindent The r should just be 1-y and the h should be 4\sqrt{1-y^{2}}. So 2\pi r h = 2\pi (1-y)\times 4\sqrt{1 - y^{2}} = 8\pi(1-y)\sqrt{1 - y^{2}}.$
Why is the radius only 1-y? Wouldn't it be 2(1-y) since 1-y only covers half the ellipse

10. ## Re: Volumes and Mechanics questions

Originally Posted by pikachu975
Why is the radius only 1-y? Wouldn't it be 2(1-y) since 1-y only covers half the ellipse
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

11. ## Re: Volumes and Mechanics questions

Originally Posted by si2136
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
Thanks heaps!

12. ## Re: Volumes and Mechanics questions

Originally Posted by InteGrand
$\noindent For the mechanics one, using v^{2} = u^{2} + 2as with v=0, u = V_{0}, s = l, we can solve for a to find that the required (constant) acceleration is a = -\frac{V_{0}^{2}}{2l}. The force is thus ma = -\frac{mV_{0}^{2}}{2l}, so the answer is (C).$
If it wasn't a multiple choice, are you allowed to use physics formulae?

13. ## Re: Volumes and Mechanics questions

Originally Posted by frog1944
If it wasn't a multiple choice, are you allowed to use physics formulae?
Generally speaking, no.

Ok, thanks

15. ## Re: Volumes and Mechanics questions

Originally Posted by frog1944
Ok, thanks
You'd use a = F/m and a = vdv/dx

16. ## Re: Volumes and Mechanics questions

Originally Posted by InteGrand
Basically the first line – it should be a = F/m. (Since F = ma.)
why isn't it a=-F/m. The force is opposing the motion

17. ## Re: Volumes and Mechanics questions

Originally Posted by tsoliman1
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.

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