Tangental acceleration from given centripetal acceleration and a range of radii

AI Thread Summary
To determine the tangential acceleration needed for a specific centripetal acceleration across various radii, the key equation is Vt^2 = r * a_c, where Vt is tangential speed, r is the radius, and a_c is the centripetal acceleration. For a centripetal acceleration of 9.81 m/s², different radii from 5 km to 100 km yield specific tangential speeds. The discussion highlights that tangential acceleration is not necessary for calculating centripetal acceleration; rather, a constant tangential speed suffices. An example illustrates that an asteroid with a 137 km circumference would require a tangential speed of 59.4 km/h to achieve the desired acceleration, suggesting potential for artificial gravity.
sneurlax
Messages
2
Reaction score
0
Hi, how can I determine the tangental acceleration of a circle needed to produce a given centripetal acceleration from a range of radii? For example, I would like to produce 9.81 m/s/s centripetal acceleration with a range of radii from 5km to 100km? All I really need is to figure out the equation and I can write a program to graphically display the results.

Here's what I've found so far:

ac = vt2/r
centripetal acceleration = (tangental acceleration)2 / radius of circular path

Fc = mvt2/r
centripetal force = mass x ((tangental speed)2 / radius of circular path)

If centripetal force is different than centripetal acceleration and a weight is needing to determine the Newtons involved then assume that the object being acted upon weighs 80 Earth kg.

Thanks for any help! If you could just nudge me in the right direction I'd appreciate it very much.
 
Physics news on Phys.org
You don't need a tangential acceleration to produce a given centripetal acceleration, all you need is a given tangential speed. This can be calculated from the first formula you gave.

Vt^2=r*a_c
where r are the different radii and a_c is the centripetal acceleration.
 
... d'oh

That seems so obvious now.

Anyways, that means that an asteroid with a 137km circumference ring drilled into it (the longest manmade tunnel so far) would need to be accelerated to a spin 59.4km/h (instantaneous velocity tangental to the ring) to produce 9.81 m/s2 acceleration... artificial gravity, anyone?
 
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top