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Mr. Heretic
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Homework Statement
EDIT: A solid homegeneous cylinder of mass M and radius R is moving on a surface with a coefficient of kinetic friction μ(k).
At t=0 the motion of the cylinder is purely translational with a velocity v(0) that is parallel to the surface and perpendicular to the central axis of the cylinder.
Determine the time t after which the cylinder performs pure rolling motion.
I've tried with energy equations to determine this and got a solution with the right units, but it seems fundamentally wrong in that for friction to do the work in converting translational to angular velocity there must be a loss of energy, right?
Will detail as "Attempt 1"
Also tried with simple kinematics and got another solution with the right units, but after doing it I realized that kinematics make all sorts of assumptions too, if mapping the trajectory of a projectile as a parabola ignores frction, then is this really any more valid?
will detail as "Attempt 2"
All thoughts and comments welcomed.
Homework Equations
E(kt) = 1/2 mv^2
E(ka) = 1/2 Iω^2
I = 1/2 MR^2
3. The attempts at a solution
Attempt 1:
E(kr i) + E(kt i) = E(kt f) + E(kr f)
0 + (1/2)mv(o)^2 = (1/2)mv(f)^2 + (1/2)Iω(f)^2
mv(o)^2 = mv(f)^2 + Iω(f)^2
No slipping at t (by definition) so v(f) = Rω(f)
mv(o)^2 = m(R^2)ω(f)^2 + Iω(f)^2
I = (1/2)mR^2
mv(o)^2 = m(R^2)ω(f)^2 + (1/2)m(R^2)ω(f)^2
v(o)^2 = (R^2)ω(f)^2 + (1/2)(R^2)ω(f)^2
v(o)^2 = (3/2)(R^2)ω(f)^2
ω(f) = √(2/3)v(o)/R
And ω(f) = ω(i) + αt
√(2/3)v(o)/R = ω(i) + αt
ω(i) = 0
t = √(2/3)v(o)/(Rα)
τ = Iα, α = τ/I
t = √(2/3)v(o)I/(Rτ)
τ = F(friction)R
F(friction) = μ(k)F(normal)
F(normal) = mg
τ = μ(k)mgR
t = √(2/3)v(o)I/(μ(k)mgR^2)
Again, I = (1/2)mR^2
t = (1/2)√(2/3)v(o)/(μ(k)g)
(1/2)√(2/3) = √(1/4)√(2/3) = √(1/6)
t = √(1/6)v(o)/(μ(k)g)Attempt 2:
v = v(i) + at
ω = ω(i) + αt
Determining variables:
v(i) = v(0)
a = F(f)/M = F(n)μ(k)/M = μ(k)g
But it's acting in the negative direction as velocity must be decreasing, so -μ(k)g
ω(i) = 0
α = τ/I
τ = F(f)R = μ(k)MgR
I = (1/2)MR^2
So α = (μ(k)MgR)/((1/2)MR^2) = 2μ(k)g/R
Plugging them in:
v = v(o) - μ(k)gt
ω = 2μ(k)gt/R
Condition of rolling: v = ωR -> v/R = ω
Equating and solving:
(v(o) - μ(k)gt)/R = 2μ(k)gt/R
v(o) - μ(k)gt = 2μ(k)gt
v(o) = 3μ(k)gt
t = v(o) / ( 3 μ(k) g )
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