An Interesting Physics Problem

[Borodog]

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_]

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]

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]

Vertical.











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

[Borodog]

[ 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]

I think it's atan(2*mu), which here is 45deg. Post a solution in a bit.

[_brady_]

[ 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]

Solid attempt, but no.

[Borodog]

[ 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_]

[ 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.