Slipping before rolling (Rotation)

[weesiang_loke]

Homework Statement

Consider a solid disc (cylinder) with mass M and radius R initially rotates with an angular velocity \omega. Then it is slowly lowered to a horizontal surface with coefficient of kinetic friction, \mu. What is the distance of the disc traveled before it starts to roll without slipping.

Homework Equations

Force, Impulse and One-dimensional kinematic equation, etc.

The Attempt at a Solution

i used -\frac{dL}{dt} = R \frac{dP}{dt}
where L is the angular momentum and P is the linear momentum.

Then i get -\Delta L = R*\DeltaP as \Deltat \rightarrow0
so, -I ( \omega_f - \omega ) = MR(\upsilon_f - 0)

after that i change the I into 0.5*M*R^2 and \upsilon_f=R*\omega_f (condition for rolling without slipping).
So my v_f = 1/3 * R *\omega.

since the frictional force is M*g*\mu, so the acceleration a = \mu*g.

After that i use the linear motion equation: v^2 = u^2 + 2as
so we have (1/3 * R *\omega)^2 = 0 + 2*(\mu*g)*s
so the distance traveled is ((R *\omega)^2) / (18*\mu*g)

The answer is correct.

But actually my question is why can we applied " -\frac{dL}{dt} = R \frac{dP}{dt} " at the beginning especially with that negative sign there. And what is the equation there stands for?

Pls help me because i am really confused here.. Thanks

 

[Hootenanny]

The equation arises from the definition of angular moment and is just a statement regarding the conservation of momentum. The definition of angular momentum for a point particle is

\mathbold{L} = \mathbold{r}\times\mathbold{P}.

Taking the derivative with respect to time yields,

\frac{d\mathbold{L}}{dt} = \frac{d\mathbold{r}}{dt}\times\mathbold{P} + \mathbold{r}\times\frac{d\mathbold{P}}{dt}.

The first term vanishes since the velocity is parallel to the momentum, leaving

\frac{d\mathbold{L}}{dt} = \mathbold{r}\times\frac{d\mathbold{P}}{dt}.

Now,for the problem in hand, the cross-product between the force (rate of change of linear momentum) and the position vector, in this case the radius of the wheel is anti-parallel to the angular momentum vector. Hence, the minus sign in your expression.

Does that help?

 

[tiny-tim]

hi weesiang_loke!

(have a mu: µ and an omega: ω )

But actually my question is why can we applied " -\frac{dL}{dt} = R \frac{dP}{dt} " at the beginning especially with that negative sign there. And what is the equation there stands for?

this is really two equations …

the rotational dL/dt = r x F, and the linear F = dP/dt …

together they make dL/dt = r x dP/dt …

in my opinion, trying to work out whether there should be a + or a - when we convert that vector equation into a scalar equation is confusing and pointless …

just say that if the cylinder originally rotates clockwise, then the friction F will be to the right, so the torque is anticlockwise, and the cylinder moves to the right …

ie L decreases while P increases

 

[weesiang_loke]

Thanks Hootenanny and tiny-tim.