Slipping transitioning to rolling without slipping

[lightofthemoon]

Homework Statement

A solid spherical ball of mass 0.75 kg and radius 5.0 cm is thrown onto a horizontal surface with coefficient of kinetic friction μ . It’s initial velocity at time t = 0 is horizontal and its initial angular velocity is zero. After rolling with slipping for a time t1 = 0.76 seconds, the ball begins to roll without slipping with angular speed ω1 = 120 rad/s. It continues to roll without slipping up a short ramp of height h = 16 cm

What is the value of μ?

Homework Equations

KE = .5mv^2 = 5Iω^2
f= μN
T=Fr=Iα
vf=vi +at

The Attempt at a Solution

I was thinking about solving this problem with conservation of energy.
.5mv^2 + .5Iω^2 - μmgd = .5mv^2 + .5Iω^2
However, we don't know the initial velocity, or the distance that the ball traveled with slipping.
In order to solve for initial velocity I thought about using one of the four kinematic equations. However, for each of the equations where are at least 2 unknowns.

So I also tried using forces to solve this problem as well.
Acceleration due to friction
f= μN
ma= μN
a=μg
Rotational acceleration
T=Fr=Iα
α= 5μg / 2r
Then I'd use the equation vf=vi +at and it's rotational equivalent. But this is where I'm stuck again, because there are also 2 unknown variables.

 

[Doc Al]

Since you're given the time, try using impulse.

 

[lightofthemoon]

I don't really understand.
Impulse= change in momentum = Ft
momentum = mv, so don't I still have to find the initial velocity as well as μ?

 

[Doc Al]

Consider angular impulse.

 

[lightofthemoon]

Oh, I see how to do the problem
But why don't I have to consider linear impulse as well since the ball is also traveling linearly?

 

[Doc Al]

Oh, I see how to do the problem

Good!

But why don't I have to consider linear impulse as well since the ball is also traveling linearly?

Since you happen to be given the initial and final angular speed of the ball, using angular impulse is the most straightforward approach. And it's all you need. (You can then calculate the initial velocity, if you like.)

 

[lightofthemoon]

Oh, ok. Thanks for your help!