The relationship between momentum and centripetal acceleration is defined by their dependence on velocity. Centripetal acceleration (a_c) is calculated as a_c = v²/R, where v is the velocity and R is the radius of curvature. Linear momentum (q) is expressed as q = mv, with m being mass. An increase in velocity affects both quantities: if velocity doubles, momentum doubles while centripetal acceleration quadruples. Thus, while they are related through velocity, they play different roles in their respective equations.
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Not sure if i get what you are asking here, but consider a point mass in motion along some curve: every single point P on the curve can be locally approximated by a circle of radius R_P (see osculating circle), so at every point you can compute a value for your centripetal acceleration a_c = \frac{v^2}{R_P}, so if you remember that linear momentum is q = mv you'll note that if velocity increase by a factor of 2, linear momentum will increase by the same factor, while centripetal acceleration will increase by a factor of 2^2=4.Sundown444 said:I see. Well, I have one more question. Centripetal Acceleration is proportional to the square of the velocity divided by the radius. How is the square of the velocity related to the velocity in momentum, if it is?
By a power of 2.Sundown444 said:How is the square of the velocity related to the velocity in momentum
mastrofoffi said:Dr.D's answer is neat, but if you're looking for something more I think you may want to clarify what you mean by being related; both of them have a dependency on velocity, yes, but that would make every cinematic quantity somehow related with each other, thus making your question a bit vague.
Not sure if i get what you are asking here, but consider a point mass in motion along some curve: every single point P on the curve can be locally approximated by a circle of radius R_P (see osculating circle), so at every point you can compute a value for your centripetal acceleration a_c = \frac{v^2}{R_P}, so if you remember that linear momentum is q = mv you'll note that if velocity increase by a factor of 2, linear momentum will increase by the same factor, while centripetal acceleration will increase by a factor of 2^2=4.
Likewise, if you change the centripetal acceleration acting on the body by a factor k (without changing its istantaneous radius of curvature), its speed, and therefore its linear momentum, will change by a factor \sqrt{k}
No, how can the velocity be independent from itself? I mean, it is the same velocity we are talking about, it just plays a different role in the 2 quantites, momentum being linear in v while centripetal acceleration is quadratic. In fact, as I showed you, with fixed mass and trajectory a change in centripetal acc implies a change in linear momentum and viceversa.Sundown444 said:I see. Well, beyond the square root of velocity in centripetal acceleration and such, there wasn't much else I was asking for, if anything at all. So, the velocity squared in centripetal acceleration is independent from the velocity in momentum, I take it?
mastrofoffi said:No, how can the velocity be independent from itself? I mean, it is the same velocity we are talking about, it just plays a different role in the 2 quantites, momentum being linear in v while centripetal acceleration is quadratic. In fact, as I showed you, with fixed mass and trajectory a change in centripetal acc implies a change in linear momentum and viceversa.
If what you are asking is "is the v^2 in the centripetal acceleration directly derivable from the linear momentum?" then for sure you can find some fancy way to derive it like that(consider a body under action of centripetal force, note that F = dq/dt, etc..) but it would be hardly meaningful with respect to the standard way of defining those quantities.