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 2010-10-28, 23:55 #1 science_man_88     "Forget I exist" Jul 2009 Dumbassville 26·131 Posts true or false could it ever be true that going down a steeper slope will get you going slower than a shallow slope for the same vertical distance. if you agree it's true can someone send it to mythbusters to test ? I don't know quite how to. though I might try to find adam savages email.
 2010-10-29, 11:30 #2 Mini-Geek Account Deleted     "Tim Sorbera" Aug 2006 San Antonio, TX USA 17·251 Posts I don't think it's possible. If you want to suggest it, you can by posting on this forum.
2010-10-29, 11:45   #3
xilman
Bamboozled!

"𒉺𒌌𒇷𒆷𒀭"
May 2003
Down not across

10,781 Posts

Quote:
 Originally Posted by science_man_88 could it ever be true that going down a steeper slope will get you going slower than a shallow slope for the same vertical distance. if you agree it's true can someone send it to mythbusters to test ? I don't know quite how to. though I might try to find adam savages email.
It is certainly possible if the two experiments are carried out in different places --- on the surface of the moon and the surface of the earth, for instance. If you want a purely terrestial experiment, immerse the steep slope in a high viscosity fluid and the shallow slope in one of low viscosity.

Paul

2010-10-29, 11:57   #4
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

838410 Posts

Quote:
 Originally Posted by Mini-Geek I don't think it's possible. If you want to suggest it, you can by posting on this forum.
apparently I forgot that link lol I've bookmarked it now lol

2011-05-04, 13:49   #5
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts

Quote:
 Originally Posted by science_man_88 could it ever be true that going down a steeper slope will get you going slower than a shallow slope for the same vertical distance. if you agree it's true can someone send it to mythbusters to test ? I don't know quite how to. though I might try to find adam savages email.
I think the reason I was thinking this is because the shallower slope although it seems counter intuitive that it would be faster it might be faster on the basis of the fact that to go down the same amount ( pull towards the center of the earth the same amount in equal time) the shallow slope has more distance to cover ( other factors include friction though).

 2011-05-04, 17:34 #6 wblipp     "William" May 2003 New Haven 1001001111102 Posts Assuming you don't want to rig the results using methods such as those suggested by Xilman above, you are converting gravitational energy into kinetic energy. If there aren't any energy leaks (friction etc), then the speed must be the same.
2011-05-04, 18:42   #7
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts

Quote:
 Originally Posted by wblipp Assuming you don't want to rig the results using methods such as those suggested by Xilman above, you are converting gravitational energy into kinetic energy. If there aren't any energy leaks (friction etc), then the speed must be the same.
13.869 angle 45
9.808 angle 89

using sine this is what I come up for for 9.80665 m drop on the opposite ( admittedly$d=v_i{t}+\frac {1}{2}at^2$ doesn't give 9.80665 m in 1 second I know).

Last fiddled with by science_man_88 on 2011-05-04 at 18:43

 2011-05-04, 19:01 #8 Zeta-Flux     May 2003 7·13·17 Posts It also depends on what you mean by slope. For example, your track could consist of two metal wires. The moving object could be two cones glued together. Then it can be the case that you can have your cones move "up" the track.
 2011-05-10, 01:48 #9 Christenson     Dec 2010 Monticello 5×359 Posts Guys: There's also an assumption of linearity here...for example, if the thing going down the slope is a rock, and at the end of the slope there's a water surface, then, if the angle is too steep, the rock falls into the water. If it is shallower, then the rock will skip, and bounce back into the air with little loss of speed. Similarly, at work we have a self-exciting brake circuit, that, basically, doesn't "catch" until a certain speed is reached. Get above that speed, you end up much slower than if you stayed just below it.
2011-05-10, 12:37   #10
R.D. Silverman

Nov 2003

22·5·373 Posts

Quote:
 Originally Posted by wblipp Assuming you don't want to rig the results using methods such as those suggested by Xilman above, you are converting gravitational energy into kinetic energy. If there aren't any energy leaks (friction etc), then the speed must be the same.
Bzzt! With respect to the original question this is false.
If the surface is frictionless and the object falls through height 'h',
then its total velocity will be 'v' as given by mgh = 1/2 mv^2 where
m is the mass and g the force of gravity.

But part of that velocity will be vertical and part of it will be horizontal.
Total velocity will be independent of slope, but VERTICAL velocity will not
be. The question was about vertical velocity.

Thank you for playing.

If the slope has angle theta (with the ground), I suggest you resolve
the velocity into its horizontal and vertical components. (I assume you
know what a dot product is??? [actually, dot products can be avoided here].

You will find that as theta --> 0 (the slope becomes flat), the VERTICAL
velocity also goes to 0...... (Think about sin(x)).

I assume of course, a frictionless slope. If it is not, then one must also
compute the frictional force (ratio of vertical and horizontal force against
the surface of contact). This will have the effect of opposing 'g', the
force of gravity. And the frictional force will depend on the slope.

This is all simple high school physics.

2011-05-10, 12:58   #11
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26·131 Posts

Quote:
 Originally Posted by R.D. Silverman Bzzt! With respect to the original question this is false. If the surface is frictionless and the object falls through height 'h', then its total velocity will be 'v' as given by mgh = 1/2 mv^2 where m is the mass and g the force of gravity. But part of that velocity will be vertical and part of it will be horizontal. Total velocity will be independent of slope, but VERTICAL velocity will not be. The question was about vertical velocity. Thank you for playing. If the slope has angle theta (with the ground), I suggest you resolve the velocity into its horizontal and vertical components. (I assume you know what a dot product is??? [actually, dot products can be avoided here]. You will find that as theta --> 0 (the slope becomes flat), the VERTICAL velocity also goes to 0...... (Think about sin(x)). I assume of course, a frictionless slope. If it is not, then one must also compute the frictional force (ratio of vertical and horizontal force against the surface of contact). This will have the effect of opposing 'g', the force of gravity. And the frictional force will depend on the slope. This is all simple high school physics.
I asked about a sloped with the same vertical height , I never remember learning this in the physics I took in high school. my argument was this assume we have a frictionless surface in that scenario to have the same vertical acceleration due to gravity the shallower slope makes the object falling have a longer distance to travel along the hypotenuse and hence have a higher velocity along the hypotenuse. with $d={v_i}t +\frac{1}{2}at^2$ at t one we have a drop of $\frac{1}{2}g$ m given this same drop the shallower slope is faster along the hypotenuse.

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