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/sci/ - Science & Math

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4345256 No.4345256 [Reply] [Original]

Say I had a ramp and I wanted to find how fast an object would accelerate down the ramp so that I could find the exit velocity what equation would I use?

>> No.4345263

0=VRoOM

V=Velocity
Ro=The vector at the origin
O= The origin itself
M= Mass

I'd highly suggest using a hotwheels car, as it will produce the most accurate results. Do NOT use matchbox cars, they suck dick.

>> No.4345267

<span class="math">\frac{d}{dt} \frac{\partial L}{\partial q'} = \frac{\partial L}{\partial q}[/spoiler]

>> No.4345292

>>4345256
conservation of energy
U = m g h

>> No.4345307

>>4345292
Yup, equate this to (1/2)mv^2 to get v at the exit

>> No.4345309

Record the height from some arbitrary point of reference, then plug into equation U = mgh where m is mass, g is 9.81 m s^-2 and h is the height you recorded. Now set that value to (1/2)mv^2 where m is your mass and v is your final velocity.
However since m is on both sides of mgh = (1/2)mv^2 you can cancel it and are left with v = square root(2gh) only problem with using this formula is that I assume your friction is 0 (which it visibly isn't) so you have to account for friction, I'm not doing that step for you. Hint: find the coefficient of kinetic friction and then just analyze the system with forces and you're good.
- BNtCG

>> No.4345318

>>4345263

I love you. I haven't had a good laugh in a while

>> No.4345324

>>4345263
Nix my way. This way even accounts for friction and everything! OP use this formula, I completely forgot about it!

>> No.4345342

>>4345267
>telling someone who clearly doesn't know any math/physics to use Lagrangian formalism where it is meaningless and tedious (friction/air resistance)

Stay silly /sci/.

>> No.4345344

<span class="math">mgh = \frac{m}{2}v^{2}[/spoiler]
Assuming friction is negligible

>> No.4345355

>>4345344
actually make that
<span class="math">gh = \frac{v^{2}}{2}[/spoiler]

>> No.4345368

>>4345355
but since you want to know the exit velocity
<span class="math"> v = \sqrt{2gh}[/spoiler]

>> No.4345405

>>4345355
but since 2g is a constant you can pull it out
<span class="math"> v =4.4295 \sqrt{h}[/spoiler]