# Fictious force: Cylinder on an Accelerating Plank

1. Feb 24, 2012

### Leb

1. The problem statement, all variables and given/known data

Problem is described in the picture

I do not understand how can $\alpha^{'}R=a^{'}$.
The dimensions do not seem correct. Angular velocity x distance from the origin = tangential velocity, is that correct ? How can this equal acceleration then ?

3. Attempt to solution

I think that $\alpha^{'}R=a^{'}$ would hold only if we would consider a unit time. That is:
$\alpha^{'}R=\frac{d\theta}{dt}R$, which, more by knowing the anticipated result in this case, than by logic, gives $\frac{d\theta}R = a^{'}dt$ which is now dimensionally OK, I think...

Last edited: Feb 24, 2012
2. Feb 24, 2012

### tiny-tim

Hi Leb!
No, a is (linear) acceleration, and α is angular acceleration

s = rθ

v = rω

a = rα

3. Feb 24, 2012

### Leb

Thanks tiny-tim !
It was strange to see an alpha instead of omega, but since the author was at times using random notation (such a ro, for distance and w for density...) made me forget about usual notation:)