A solid disk is given an initial angular speed of 5.5 rad/s and then lowered to a table. The disk skids for a time, but then achieves pure rolling. What is the speed of the center of mass during pure rolling?

I know the equation for this is Vcm = angular velocity * radius, But I couldn't figure out how to find the radius given angular velocity only

Question

A solid disk is given an initial angular speed of 5.5 rad/s and then lowered to a table. The disk skids for a time, but then achieves pure rolling. What is the speed of the center of mass during pure rolling?

I know the equation for this is Vcm = angular velocity * radius, But I couldn't figure out how to find the radius given angular velocity only

irishboy123 · Answer

draw it

mrnood · Answer

If it skids for a time then it will lose energy and hence angular velocity. We don't know how much.
If it is 'frictionless' then it will never roll

this question is impossible to solve sensibly.

Even if you incorrectly assume that angular velocity reamins the same then I still think you need to know r (as you suggested)

icrazyaliens · Answer

Easy, but I am not sure about my answer.

The kinetic energy of rotational is 
$$K = (1/2) I w^2$$
and the potential energy is 
$$P = mgR$$
The rotational inertia for the disk is 
$$I = (1/2)mR^2$$
the angular speed is 
$$w = v/R$$

Now equals them together 
$$mgR = 1/2 (1/2mR^2)w^2$$
solve for R 
$$R = (4g/w^2)$$
once you find R 

get the velocity by this formula 
$$v = wR$$

mrnood · Answer

I don't think this is a valid approach - the potential energy of the disk is not relevant as far as I can see.

You COULD argue that the KE of the spinning disk = combined KE spinning and translating

but you don't know I or R so cannot equate those things
(not forgetting the 'lost energy' through the slipping..)

vincent-lyon.fr · Answer

You need to know the radius of the disc and the kinetic friction coefficient.
It's quite an advanced problem in motion of solid bodies.

daved1948 · Answer

Skip the unnecessary details. It will lose exactly one half it's rotational energy accelerating itself with respect to the surface up to the point where skidding ceases regardless of it's mass or the coefficient of friction. The answer is (2.25 r). There is no numerical answer for this question without knowing r.

mrnood · Answer

@daved1948 
please show your derivation of this

@Vincent-Lyon.Fr and I are competent physicists and I cannot see how this can be solved.

your assertion is not sufficient

vincent-lyon.fr · Answer

Actually, for a solid disc, 2/3 of the energy is lost in friction and 1/3 is retained in total kinetic energy.
1/9 is rotational KE, and 2/9 is translational KE.

irishboy123 · Answer

LOL!!

mrnood · Answer

Anyone going to show ther derivation of these assertions?

We have a disk of unknown mass and radius and we have a surface of unknown friction.

We know just the angular velocity at t0.

I would love to see the proof that we can derive any solution for the rolling state.....

irishboy123 · Answer

|dw:1479982195883:dw|


In terms of rotation, the torque from the friction is 

$\tau = \mu W r = - I \ddot \theta$

$\mu mg r = - \dfrac{1}{2} mr^2 \ddot \theta$

$\ddot \theta = - \dfrac{2 \mu g}{r}  $

$\dot \theta = - \dfrac{2 \mu g}{r}  t + \dot \theta_o$

In terms of linear motion

$F = \mu mg = m \ddot x$

$\ddot x = \mu g$

$\dot x = \mu g t$

slipping stops when $\dot x = r \dot \theta$

$ \mu g t =  r ( - \dfrac{2 \mu g}{r}  t + \dot \theta_o )$

$  t =  \dfrac{r \dot \theta_o}{3 \mu g} $

At that point:

$\dot x = \mu g \cdot \dfrac{r \dot \theta_o}{3 \mu g} $

$= \dfrac{5.5r}{3}$