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.

 

[rwisz]

The equation you used, -\frac{dL}{dt} = R \frac{dP}{dt}, is known as the Euler's equation of motion. It is derived from the principle of conservation of angular momentum, which states that the total angular momentum of a system remains constant unless acted upon by an external torque. In this case, the external torque is the frictional force acting on the disc.

The negative sign in the equation represents the direction of the torque and is necessary to account for the change in direction of the angular momentum. Without the negative sign, the equation would not be able to accurately predict the behavior of the disc as it slows down and starts rolling without slipping.

Essentially, the equation is a mathematical representation of the physical concept that for every action (change in linear momentum), there is an equal and opposite reaction (change in angular momentum). It allows us to analyze the motion of the disc in terms of both linear and angular quantities, and ultimately leads to the correct solution for the distance traveled before the disc starts rolling without slipping.

I hope this helps clarify the use of the equation in this problem. If you have any further questions, please let me know.