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-   -   An Interesting Physics Problem (http://archives1.twoplustwo.com/showthread.php?t=547262)

Borodog 11-15-2007 11:22 PM

An Interesting Physics Problem
 
What is the maximum slope that a solid sphere of uniform density can roll down without slipping, if the coefficient of static friction between the sphere and the incline is 1/2?

_brady_ 11-15-2007 11:41 PM

Re: An Interesting Physics Problem
 
26.6 degrees (from the horizontal)?

That can't be it or else this wouldn't be interesting...I give up.

Edit - Can I have multiple answers? I want to add an answer of 63.4 degrees from the horizontal.

Borodog 11-16-2007 01:21 AM

Re: An Interesting Physics Problem
 
Neither of those two answers appear to be correct.

Perhaps if you provided a worked out solution.

Edit: Ah, never mind. I see how you got that.

Definitely not.

Arp220 11-16-2007 01:50 AM

Re: An Interesting Physics Problem
 

Vertical.











(you didn't say I couldnt add some, uuummmm.... prespin [img]/images/graemlins/smile.gif[/img] )

Borodog 11-16-2007 01:58 AM

Re: An Interesting Physics Problem
 
[ QUOTE ]

Vertical.


[/ QUOTE ]

Wrong.









[ QUOTE ]

(you didn't say I couldnt add some, uuummmm.... prespin [img]/images/graemlins/smile.gif[/img] )

[/ QUOTE ]

Prespin all you want. It will still slip. Do you see why?

Andy Ross 11-16-2007 07:15 AM

Re: An Interesting Physics Problem
 
I think it's atan(2*mu), which here is 45deg. Post a solution in a bit.

_brady_ 11-16-2007 10:20 AM

Re: An Interesting Physics Problem
 
[ QUOTE ]
Neither of those two answers appear to be correct.

Perhaps if you provided a worked out solution.

Edit: Ah, never mind. I see how you got that.

Definitely not.


[/ QUOTE ]

Ya, the first one I was sure was wrong and I just realized I did the second one wrong, so here is a reworked solution.

x direction parallel to slope, y direction perpendicular, angle is theta

Sum moments bout point of contact, P:

mgrsin(theta)=Ip*alpha

Ip=2/5mr²+mr²

d²x/dt² = a_x = r/2*alpha

solveing for acceleartion in x direction a_x

a_x = (5/14)*g*sin(theta)

Now summing forces in x direction:

-Ff + mgsin(theta) = m*a_x = m*(5/14)*g*sin(theta)

where Ff = mu*N = 0.5*N = 0.5mg*cos(theta)

-0.5mgcos(theta)+mgsin(theta)=5/14*mg*sin(theta)

solving for theta:

theta = 37.9 degrees from the horizontal.

Borodog 11-16-2007 11:43 AM

Re: An Interesting Physics Problem
 
Solid attempt, but no.

Borodog 11-16-2007 11:46 AM

Re: An Interesting Physics Problem
 
[ QUOTE ]
I think it's atan(2*mu), which here is 45deg. Post a solution in a bit.

[/ QUOTE ]

Nope.

Let me just give the numerical answer so you'll know when you've gotten it.

The slope is 7/4.

_brady_ 11-16-2007 03:21 PM

Re: An Interesting Physics Problem
 
[ QUOTE ]
[ QUOTE ]
I think it's atan(2*mu), which here is 45deg. Post a solution in a bit.

[/ QUOTE ]

Nope.

Let me just give the numerical answer so you'll know when you've gotten it.

The slope is 7/4.

[/ QUOTE ]

Oops I think I made one other silly mistake. I get this if I modify one line in my solution:

Instead of:
d²x/dt² = a_x = r/2*alpha

If I use:
d²x/dt² = a_x = r*alpha

I get a slope of 7/4 or angle of 60.3 deg.


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