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*$$\Delta$$P as $$\Delta$$t $$\rightarrow$$0
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.