Fictious force: Cylinder on an Accelerating Plank

[Leb]

Homework Statement

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...

 

[tiny-tim]

Hi Leb!

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 ?.

No, a is (linear) acceleration, and α is angular acceleration

s = rθ

v = rω

a = rα ​

 

[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:)