Rolling and Sliding: Solving Angular Momentum Theorem

[roadrunner1994]

Homework Statement

A homogeneous ball with mass M and radius R is set in
motion on a horizontal floor in such a way that the ball initially (i.e., at
time t = 0) has the translational velocity Vo. We assume that the ball begins
its motion as pure sliding. The coefficient of friction between the ball and
the floor is μ. The motion of the ball between the time t = 0 and t = tr is
a combination of sliding and rolling. At time t = tr the ball begins to roll
without sliding. In the time between t = 0 and t = tr the ball has moved a
distance D along the floor.
(1) Determine tr.

Relevant Equations

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Hi,
I solved this prolem in the following way.
I have started with the angular momentum theorem:
Iα=fR
As the force of friction vector is attached on point of contact of the ball and the supporting surface, the moment of inetria is:
Ic= MR^2 + Icm = MR^2 + (2/5)MR^2

Ic*a/R=μMgR
tr=7/5(v0/μg)

It seems that I've made a mistake during calculations of moment of inertia, because the right anwer is tr=2/7(v0/μg), but I can't find that mistake.
Thanks for your help!

 

[phyzguy]

The ball's velocity is slowing down due to the force of friction. It seems that you haven't taken that into account, have you?

 

[BvU]

force of friction vector is attached on point of contact of the ball and the supporting surface

That point of contact is not a fixed axis of rotation

 

[roadrunner1994]

You are right, I think the solution will be:

a = -μg
α = (5/2)μg/R

v = v0 - μgt
ω = (5/2)μgt/R

and at the moment when ball starts to roll
v0 - μgt = R* (5/2)μgt/R
t = 2/7(v0/μg)

 

[roadrunner1994]

One more question
(3) Determine the work W that the forces of friction perform in the time
from t = 0 to t = tr i.e. in the time where the ball was both rolling and
sliding

I suppose I can't just use the equation _t0∫^trFvdt, can I?

 

[Hamiltonian]

One more question
(3) Determine the work W that the forces of friction perform in the time
from t = 0 to t = tr i.e. in the time where the ball was both rolling and
sliding

I suppose I can't just use the equation _t0∫^trFvdt, can I?

you could simply apply the work-energy theorem or since the force of friction remains constant till time tr you could just use $W = f.r$ where r is the displacement of the ball.

 

[roadrunner1994]

you could simply apply the work-energy theorem or since the force of friction remains constant till time tr you could just use $W = f.r$ where r is the displacement of the ball.

In this way? W = ΔK = ½Mv^2 + ½Iω^2 - ½M(v0)^2
since
v = v0 - μgt
ω = (5/2)μgt/R
t = 2/7(v0/μg)

the result is negative → W = -1/7[M(v0)^2]
Is it correct?

 

[phyzguy]

You subtracted the initial kinetic energy from the final kinetic energy. It comes out negative because kinetic energy has been lost to friction and converted to heat.

 

[roadrunner1994]

Thank you
And for the subsequent subpoint:
Show that W = ₀∫^trμMgVc(t)dt, where Vc is the velocity of the contact
point relative to the supporting surface (the floor).

The velocity of the contact point will be the following:
Vc = ωR - v = ωR - v0 + μgt, right?