Fictious force: Cylinder on an Accelerating Plank

In summary, the conversation discusses a problem involving the equation α'R=a', which does not seem to have correct dimensions. It is suggested that this equation would only hold if a unit time was considered and that knowing the anticipated result in this case would make it dimensionally correct. The conversation also clarifies that α is angular acceleration and not to be confused with ω, which represents linear acceleration.
  • #1
Leb
94
0

Homework Statement



Problem is described in the picture
Cylinder on an accelerating plank.jpg

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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  • #2
Hi Leb! :smile:
Leb said:
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 :wink:
 
  • #3
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:)
 

1. What is a fictitious force?

A fictitious force is a perceived force that appears to act on an object due to its motion in a non-inertial frame of reference. It is not a real force, but rather a mathematical correction used to explain the motion of objects in non-inertial frames.

2. How does the cylinder on an accelerating plank experience a fictitious force?

The cylinder on the accelerating plank experiences a fictitious force, also known as the centrifugal force, because it is moving in a circular motion in a non-inertial frame of reference. This perceived force is not actually acting on the cylinder, but is necessary to account for its motion in the non-inertial frame.

3. How is the magnitude of the fictitious force on the cylinder determined?

The magnitude of the fictitious force on the cylinder is determined by the mass of the cylinder, its velocity, and the radius of its circular motion. This can be calculated using Newton's second law, where the fictitious force equals the mass of the object multiplied by its centripetal acceleration.

4. Can a fictitious force be felt by an object?

No, a fictitious force cannot be felt by an object because it is not a real force. It is simply a mathematical correction used to explain the motion of objects in non-inertial frames of reference.

5. How does the acceleration of the plank affect the fictitious force on the cylinder?

The acceleration of the plank does not affect the fictitious force on the cylinder, as long as the cylinder maintains a constant velocity in the circular motion. The magnitude of the fictitious force remains the same, but the direction may change depending on the direction of the acceleration.

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