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Fictious force: Cylinder on an Accelerating Plank

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Leb
#1
Feb24-12, 03:53 PM
P: 94
1. The problem statement, all variables and given/known data

Problem is described in the picture
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I do not understand how can [itex]\alpha^{'}R=a^{'}[/itex].
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 [itex]\alpha^{'}R=a^{'}[/itex] would hold only if we would consider a unit time. That is:
[itex]\alpha^{'}R=\frac{d\theta}{dt}R[/itex], which, more by knowing the anticipated result in this case, than by logic, gives [itex]\frac{d\theta}R = a^{'}dt[/itex] which is now dimensionally OK, I think...
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tiny-tim
#2
Feb24-12, 04:42 PM
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tiny-tim's Avatar
P: 26,148
Hi Leb!
Quote Quote by Leb View Post
I do not understand how can [itex]\alpha^{'}R=a^{'}[/itex].
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
#3
Feb24-12, 05:03 PM
P: 94
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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