help on a physics question...

[chiwawa626]

Consider this situation : A long cylindrical candle (diameter of at least 2 cm) has a needle stuck through it's center of mass. The candle is suspended by this needle, which is sitting on two objects, so that the candle is free to pivot about this axis created by the needle. In this initial suspended position, the candle is approximately horizontal.
IF you light both ends of the candle and allow it to burn,
THEN it will oscillate with a frequency inversely proportional to its length as it burns.
Why is there a higher frequency as the length gets smaller?

 

[Viper0329]

Beats me

 

[Garlicboy]

Less mass

 

[glen]

Once you fix the center of mass of the candle, it has 2 new centers of mass on either side of the pin.
As the candle burns, these 2 new centers of mass move closer and closer to the pin.
Look at your formulas for circular motion.

 

[chiwawa626]

Glen, but, Why is there a higher frequency as the length gets smaller? and also why does it occilate?

does it occilate beacuse the burning of the string causes an inbalance in the 2nd center of mass (1st center is where the pin is...) ?

 

[glen]

r is your radius which is the distance from the pin

V=2pi r / t

F=ma=mv^2/r

 

[glen]

Not exactly sure what your professor is asking, but it should be clear that a pendulum oscillates faster as the string becomes shorter...same thing when the candle becomes shorter.

 

[chiwawa626]

bump anyone else can answer this? please help i need the credit for this

 

[helikon]

My guess is that if the forces causing the oscillation (vibrations due to combustion process...gas dynamics anyone?) are essentially constant for the duration,then as the candle gets shorter it's mass moment of inertia about the axis of rotation decreases.So instead of F=ma it's more like T=I*a (that a is supposed to an "alpha").Like someone said earlier,less mass. helikon

 

[Mangos]

Nevermind. It's not doing full circles. Sorry.

 

[Mangos]

am I right?

No, i'm not. Sorry.

 

[merlocka]

Assuming the oscillation occurs because one side will evaporate more wax as it is angled downward from horizontal (and therefore become lighter... hence the oscillation), the frequency will increase because there will be less and less of a moment arm to rotate.

My Tipler physics book is way too buried to pull out but it shouldn't be that bad to equate.

 

[doublec16]

The torque on each end of the candle is given by I(alpha) where I is the moment of inertia (proportional to Mr^2, go ahead and look up the moment of inertia of a cylinder with the axis at its center) and alpha is the angular acceleration. Adding the torques cancels them out, producing an equilibrium. Now displace the candle by a small angle. The period of oscillation is proportional to the square root of I. I might be missing something here, but since the mass is proportional to the length, assuming uniform density and width, which characterizes the candle you describe, the moment of inertia is proportional to r^3, and therefore the frequency is proportional to r^(-3/2), and therefore also length^(-3/2) and not inverse length as the problem suggests. Anyway, since the exponent is negative, the frequency will increase with decreasing length.

 

[RossGr]

Seems to me that the burn rate will vary if the end is up or down, I think the lower end will burn/melt faster, but that means it will become shorter/lighter then ossiclate to the upper postion, where upon the other ends starts burning faster. Thus the osillations.

EDIT:
Opps, sorry, just saw merlocka's post, he beat me to it! The moment of inertia post above may well be exatly correct, sorry a bit hard to follow. Think of it like a form of simple pendulum where period of ossilation goes like the lenght. since period in inverse of frequecy the frequency is inverly porportional to lenght.