Question

Q. If an object of mass m is moving uniformly (constant speed v) in a circle of radius r:

1.State the equation that relates the centripetal force, Fcent, to m, v, and r.

A: my answer, Fcent= mv^2/r

2.State the equation that relates the period for one revolution, T, to v and r.

A:My answer, T=2pir/v

3.From the two previous equations write an equation relating T, to m and r (show your working0

A: I don't know how to do this part ????

Answer 1

Solve each equation for v, and then equate those two equations.

Answer 2

thanks, I calculated v=\[v=\sqrt{fmr}/m\] and \[v=2\pi r/T\] , I then solved to get

T=\[2\pi \rm/\sqrt{fmr}\]

but i don't want to have F in there as it says to relate T to m and r and not F. are my calculations correct? how do i get rid of the F

Answer 3

First, recheck your math, specifically your first equation. As for the force, you can't get rid of it. Force and velocity vary together. Since period is a function of velocity, then period also has to vary with force

Answer 4

Thanks, yeah i realised i made a mistake with my first equation, it should be +-sqrt(fr/m), which gives me \[T=2\pi r \sqrt{m}/\sqrt{fr}\], does this look better?

Answer 5

\[Equation 1: F=\frac{ mv ^{2} }{ r }\]\[Equation 2: T=\frac{ 2\pi r }{ v }\]\[Equation 3: Solve Eqn 2 for v: v=\frac{ 2\pi r }{ T }\]Now insert Eqn 3 into Eqn 1:\[F=\frac{ 4\pi ^{2}mr ^{2} }{ T ^{2} r }=\frac{ 4\pi ^{2} mr}{ T ^{2} }\]\[T ^{2}=\frac{ 4\pi ^{2} mr}{ F }\]\[T=2\pi \sqrt{\frac{ mr }{ F }}\]

Answer 6

Now that I look at it, the only difference is that my answer is simplified, while your's isn't given you've got an r in the numerator and the square root of r in the denominator. So your's is correct in terms of units but is just not simplified.

Answer 7

Note that you don't have to worry about the ± because period is always positive.