Time needed to start rolling without slipping?

Question

A uniform spherical shell of mass M and radius R, spins about its axis with
an angular velocity ω. The cylinder is now dropped on a horizontal table located
slightly below the cylinder. The coefficient of friction between the cylinder and the
table top is μ. The cylinder then initially slips relative to the table, but after a time T
it continues to roll without further slipping. Express your answers to the following
questions in terms one or more of the known quantities. M, R, ω0 and μ.
a) How large is the time T?
b) What is the speed of the spherical shell’s center at the instant when the spherical
shell stops slipping?
c) Suppose that the spherical shell had twice as large a radius. If the spherical shell is
made of the same material, its mass would then be four times as large. How much
larger would be the time T? How much larger would be the final speed of the
spherical shell’s center?

Answer 1

I didn't like my first answer, so I re-thought and tried a different approach. I made it way, way too hard before!

If we consider the linear acceleration of the sphere after contacting the table, that is
M*g*μ=M*a
a is constant until the sphere stops slipping
since the transition from kinetic friction slipping to static friction, no slipping is non-linear.
so a=g*μ
using the equation of motion for the translational velocity of the sphere
v=a*T
or
v=g*μ*T
now, when the sphere doesn't slip, v=ω*R
so
ω=g*μ*T/R

with me so far?

Okay, now
while the sphere slips, the torque due to friction is
M*g*μ*R
this is equal to I*alpha
or alpha=g*μ*M*R/I
since I=2/5*M*R^2
alpha=5*g*μ/(2*R)
using the rotational equation of motion
ω=ω0-alpha*T
or
ω=ω0-5*g*μ*T/(2*R)
still with me?
See how the relationship between v and ω related the two frames of reference?

so
ω0-5*g*μ*T/(2*R)=g*μ*T/R
T*g*μ*(1/R+5/(2*R))=ω0 or
T*g*μ*7/(2*R)=ω0
T=ω0*2*R/(g*μ*7)

Again, you can compute v from T. You can also see how T and V are affected by R

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