/sci/ - Science & Math » Thread #11532150

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Anonymous Sat 04 Apr 14:35:38 2020 No.11532150 View Reply Original Report

Quoted By: >>11532157 >>11532203 >>11532517 >>11532868

Lets do some saturday morning recreational math

A ball (sphere/disk/ring/pointmass, pick your own difficulty level) rolls off an edge
How does its horizontal velocity increase?

Anonymous

Anonymous Sat 04 Apr 2020 14:38:09 No.11532157 Report

Quoted By:

>>11532150
not

Anonymous

Anonymous Sat 04 Apr 2020 14:56:56 No.11532189 Report

Quoted By: >>11532384 >>11532862

If the ball is moving slowly enough, it should get a push from the corner. Consider the limiting case of a motionless ball starting out tangent to the corner at a 45 degree angle. It can't fall vertically. It won't stay where it is. It has to accelerate a bit to the right.

Anonymous

Anonymous Sat 04 Apr 2020 15:00:53 No.11532203 Report

Quoted By: >>11532233

>>11532150
is it rolling or sliding?

Anonymous

Anonymous Sat 04 Apr 2020 15:08:14 No.11532233 Report

Quoted By:

>>11532203
if you dont want to deal with rotational inertia then assume sliding.
i found it easier to do that first as a sanity check and then redo the parts where rotation matters

Anonymous

Anonymous Sat 04 Apr 2020 16:05:04 No.11532384 Report

Quoted By:

>>11532189
>Consider the limiting case of a motionless ball starting out tangent to the corner at a 45 degree angle. It can't fall vertically. It won't stay where it is.
If there is no friction like in all physics problems then it will indeed stay where it is.

Anonymous

Anonymous Sat 04 Apr 2020 16:12:47 No.11532407 Report

Quoted By:

I'll start with an easier problem, for the critical velocity past which the disk/ring/sphere's horizontal velocity doesn't change. Parameterize the disk's bottom with
$y=-\sqrt{R^2-x^2}\\<br />y''=\frac{R^2}{(R^2-x^2)^{3/2}}$
So the vertical curvature at x=0 is 1/R, and it strictly increases with greater x. This is useful since gravity produces constant vertical curvature of the ground from the ball's reference frame. So the ball's velocity needs to be high enough to produce vertical spatial curvature of the ground of less than 1/R.
For the ground,
$x=vt\\<br />y=gt^2/2=gx^2/(2v^2)\\<br />y''=g/v^2=1/R\\<br />v=\sqrt{gR}$

Anonymous

Anonymous Sat 04 Apr 2020 16:18:00 No.11532417 Report

Quoted By:

I'll start with an easier problem: for the critical velocity past which the disk/sphere/ring's horizontal velocity doesn't change. Parameterize the ball's bottom with
$y=-\sqrt{R^2-x^2}\\ y''=\frac{R^2}{(R^2-x^2)^{3/2}}$
So the vertical curvature at x=0 is 1/R, and it strictly increases with greater x. This is useful since gravity produces constant vertical curvature of the ground from the ball's reference frame. So the ball's velocity needs to be high enough to produce vertical spatial curvature of the ground of less than 1/R. For the ground,
$x=vt\\ y=gt^2/2=gx^2/(2v^2)\\ y''=g/v^2=1/R\\ v=\sqrt{gR}$

Anonymous

Anonymous Sat 04 Apr 2020 17:12:30 No.11532517 Report

Quoted By: >>11532803

>>11532150
it does not.

Anonymous

Anonymous Sat 04 Apr 2020 19:10:43 No.11532803 Report

Quoted By:

>>11532517
Depends on whether the collision between the ball and the surface was elastic/inelastic, also depends on the friction coefficient. Or am I brainlet?

Anonymous

Anonymous Sat 04 Apr 2020 19:25:50 No.11532858 Report

Quoted By: >>11533065

assumptions
>the surface is smooth
>the surface is massless
>this occurs in deep space, with negligible gravitational effect from celestial bodies and no air resistance/drag
>the ball is sliding, not rolling

$vx,1 = vx,0$

Anonymous

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Anonymous Sat 04 Apr 2020 19:27:41 No.11532862 Report

Quoted By:

>>11532189
>tangent to the corner

Anonymous

Anonymous Sat 04 Apr 2020 19:29:33 No.11532868 Report

Quoted By:

>>11532150
Roll off edge bounce once continues to roll

Anonymous

Anonymous Sat 04 Apr 2020 20:29:07 No.11533065 Report

Quoted By:

>>11532858
nice

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