TBR pg 101: Centripedal acceleration

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Question 9 on Pg 110 passage II of chapter 2 in TBR.

Three identiical bugs (marked A B and C) are standing on a turntable as it begins to spin. as the spinning increases, which of the bugs is most likely to slip first?

(PICTURE of a circle, with radius/center marked. A is close to the center, then B and then C which is closest to the edge of the circle)

The anwer is Bug C is likely to slip first b.c its furthur from the center.
I know that F = m* v^2 / r so bug A with the smallest r (distance) will experience the greatest force.

kindly help explain

 

[Balantidiumcoli]

The formula is right but its application is wrong.
F = m* v^2 / r

Let me make a fake circle of r = 10m
Assume a rotation takes 1 second

The inner most part of the circle, d = 0 m from center
The outer most part of the circle, d = 10 m from center

v = d/t

Now, with respect to a rotation, these spots on the disk all complete a rotation at the same time.

*BUT*

The distance travelled by the bugs at these spots during a single rotation varies significantly (and hence the velocity changes as well). A bug at the 10 m radius travels a distance of 20pi m (v = 20pi m/s) and the inner most bug travels a distance of nearly 0 (v = 0). Apply the equation above:

F1 (bug farthest away) = m * ((20pi)^2) / 10 = m*40(pi^2)
F2 (nearest bug) = 0

** Circumference = 2piR **

 

[FeinMS]

Another form of the centripetal force is F=m(w^2)r. This comes from the fact that V=wr and substituting wr for V in mv^2/r gives that equation. In the equation F=mv^2/r, notice that not only r is different but their tangential velocity (v) is also different, so you can't make a valid comparison. However, in F=m(w^2)r, angular velocity (w) of the rotating disc is the same at all points on the disc, so you can make a comparison solely based on r. Notice F increases as r increases. Therefore, the outermost bug experiences the most force and most likely to slip.