I think I get what you're trying to say, but I don't believe it's quite right. First, your 'r' term comes from an equation for torque = moment of inertia times angular acceleration:
$\displaystyle T = I \alpha $
For a solid sphere rolling on an incline it may seem that $\displaystyle T = mg \sin (\theta) R, \ \ I = (2/5)mR^2$ and [math] \alpha = \frac a R[/math, which yields:
$\displaystyle a = \frac 5 2 g \sin (\theta) $
which is what you added to your "e" sliding acceleration term. The problem with this is that it ignores the fact that the ball must accelerate in translation as well as rotation. You can't simply add this to the acceleration you calculated for a sliding object - in effect this is saying that a rolling object can accelerate faster than a sliding object, even if there is no friction, which is clearly not right. Note also that if sin(theta) = 1 this would say that the ball accelerates at a rate faster than g, which again is not right.
A better way is to apply energy methods:
$\displaystyle \Delta PE = - \Delta KE $
For the rolling ball on an incline:
$\displaystyle mgd \sin(\theta) = \frac 1 2 mv^2 + \frac 1 2 I \omega ^2 $
where 'd' = displacement along the incline. If you work it though you should find that:
$\displaystyle v = \sqrt {\frac {10} 7 gd \sin \theta} $
Note that the friction force does not contribute to the analysis, as long as the ball is rolling and not sliding. Also note that this assume that $\displaystyle v_0 = 0$. You can then determine distance as a function of time:
$\displaystyle v = \frac {dx}{dt} = \sqrt { \frac {10} 7 g \sin \theta x}$
(Note that here I have switched from 'd' to 'x' for the distance the ball has rolled down the ramp.)
$\displaystyle \int \left( \sqrt {\frac {10} 7 g \sin \theta}\ \right) dt= \int x ^{-1/2} dx $
$\displaystyle \left( \sqrt {\frac {10} 7 g \sin \theta}\ \right) t = 2 \sqrt x $
$\displaystyle x = \frac 5 {14} g \sin \theta t^2 $
One other point: all this assumes the ball is rolling down the incline. But if the friction force is small enough there isn't enough torque provided by friction to get the ball rolling and it will slide rather than roll. So we need to check whether that is the case. In order to roll, the torque from friction must be at least equal to the torque required to get the ball spinning at the required rate:
$\displaystyle T = \mu mg \cos \theta \ge I \alpha$
We can find alpha from the second derivative of the equations we derived above: $\displaystyle a = \frac {d^2x}{dt^2} = \frac 5 7 g \sin \theta$, and $\displaystyle \alpha = \frac a {r}$. Plug this in, along with the known value for I:
$\displaystyle \mu mg \cos \theta r \ge \frac 2 5 mr^2 \frac 5 7 \frac {g \sin \theta} r $
which leads to:
$\displaystyle \mu \ge \frac 2 7 \tan \theta$
If mu is less than this value, the ball will slide while slowly beginning to rotate, which complicates the analysis. Post back if you want to go deeper into what happens under this condition.
Hope this helps.
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Last edited by ChipB; Mar 13th 2017 at 03:17 PM.
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