Tsao's Questions (sorry if they are too noob >< )

[astarman]

umm well ill start off with the prac sac questions.. its a uniform circular motion prac...

1) derive the relation between the speed V , the radius of the circle R, and the frquency of revolutions f   ( is that just station the formula? :S )

2) relationship between F (force ) and V ? is that just F is proportional to V ?

3) suppose that the relation between F and f is the form F=kf^n , where k and n are constants. by taking the "log" of both sides of the expressions, write the equation with log F as the subject? er... logF = n Log kf ?? lol =='  .... what will be the shape of the graph of logF against logf... and what would the gradient of such graph represent ?

sorry if the questions are too noob, but im really noob lol.. well any help will be greatly appreciated ,
thanks Alex

[kyzoo]

1.)

Speed = Distance/Time
Velocity = Distance/Time = Circumference of Circle/Period, period is the amount of time needed to make a full rotation

Because Frequency = 1/Period

Therefore Velocity = Frequency x Circumference of Circle
V = f x 2piR

2.)

F = v²/r

Therefore F is proportional to V²


3.)

What is F and what is f?

[astarman]

F is force, and f is frequency, sorry , forgot to state it

oh yea.. THANKS for the help... really appreciate it

[appianway]

3)

Put both sides in log(10) or ln

So, ln(F) = ln(kf^n)

As ln (ab) = ln(a) + ln (b), you can separate the two

So, ln (F) = ln(k) + ln(f^n)
= ln(k) + nln(f)

A graph of ln (F) vs ln (f) will thus give a linear graph with a gradient of n with a y intercept of ln(k) (neglecting the influence of systematic errors).

[astarman]

oh... i see... thanks, so wen they mean wat the gradient represents , they dnt mean a actual thing... lol =='
couldnt of done it without you guys.. thanks a lot

[appianway]

No problem! I love anything relating to labs.

Think of the equation in terms of a linear equation: you have a variable term and another constant, and the variable term is multiplied by a constant (giving you a gradient). It doesn't matter that you have a log graph.