Fictious force: Cylinder on an Accelerating Plank

by Leb
Tags: accelerating, cylinder, fictious, force, plank
 P: 88 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...
HW Helper
Thanks
P: 26,167
Hi Leb!
 Quote by 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
 P: 88 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:)

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