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Solid uniform cylinder rolls down the plane inclined
45 degrees to the horizon. Coefficient of friction
between the cylinder and the plane is μ.

What value of friction μ results in greatest possible
loss of energy due to friction?

2007-05-21 07:06:12 · 1 answers · asked by Alexander 6 in Science & Mathematics Physics

"m*g*µ*L*sqrt(2)/2"

The work done by the force of friction
is not done over the distance L, which
cylinder travelled, but rather over how
much the surface of the cylinder slipped
wrt the inclined plane.

2007-05-22 04:18:15 · update #1

1 answers

I will assume that the µ is kinetic friction.
I will also note that the incline length will be assumed to be infinite, so that an infinite amount of frictional work is done (I guess this is a bit out there since the length would have to be short enough to keep the assumption that g is constant and the cylinder is near the surface of the Earth). No matter, the maximum friction is done when µ is infinitesimally smaller than .5. Under this condition the cylinder will slip and the angular rotation will never catch up to the translational since the translational is
v(center of mass)=t*g*(1-µ)*sqrt(2)/2

and the rotational speed of the edge of the cylinder is
v(edge)=t*µ*g*sqrt(2)/2

note that for any t>0 and µ<.5
v(center of mass) so the thing keeps doing work as long as the incline presents additional length. The µ must be infinitesimally close to .5 to get maximum work since the work of friction will be m*g*µ*L*sqrt(2)/2 for any L. So the bigger the µ, the more work that gets done.
The work done over the distance traveled by the edge of cylinder is related to the distance traveled by the center of mass, the rotational distance by the edge is
L*µ/(1-*µ)


BTW: This is generally true for µ<1/tan(th)

Looking at energy, there is the potential energy lost, which is
m*g*L*sqrt(2)/2
The gain in kinetic energy of the center of mass
.5*m*v^2
the gain in rotational kinetic energy
.5*I*w^2
the loss due to friction in the translational
m*g*L*μ*sqrt(2)/2
the loss due the frictional in the rotational
m*g*μ*R*th, where th is the total rotation over the length L

j

2007-05-21 13:19:13 · answer #1 · answered by odu83 7 · 0 0

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