Problem proving tension in string

[Dracovich]

Ok so i have a homework problem again :) This time it's a simple pendulum swing and the question is as follows:

"Prove that the tension in the string of a simple pendulum when the string makes an angle \theta with the vertical is approximately mg(1 + \theta^2_0 - (3/2)\theta^2 where m is the mass of the bob and \theta_0 is the value of \theta at the extremes of the motion."

Now our teacher gave us a bit of help, saying that we should use a taylor polynom to solve this, as well as the formula:

x(t) = x_0 * cos(\omega * t)

Where x_0 is x (or s if you prefer) at the extremes of it's motion.

So i basicly want to find the formula for the total force on the string right? So i got the resultating forces:

F_{res} = mg*cos(\theta) + F_{string} = ma = m(v^2/r)

And hence i isolate the force on the string

m(v^2/r) - mg*cos(\theta) = F_{string}

I'm not entirely sure what to do next. I'm not sure if i should be making the Taylor Polynom out of

x(t) = x_0 * cos(\omega * t)

Can i simplify that down to only cos(theta) ? That would be easy enough and could perhaps get that to fit the formula given in the question (i'm thinking maybe if i could do that i could replace v^2 in the above example with 2Ax and use the taylor polynom insted of x in the point theta_0 and then substitute cos(theta) with the taylor polynom of theta). Someone also tried to hint to me i should use the preservation of energy, but i'm not seeing where that should enter the picture.

If someone could perhaps nudge me in the right direction that would be great :o (now i just gotta hope all that latex came through as intended).

[Doc Al]

So i basicly want to find the formula for the total force on the string right? So i got the resultating forces:

F_{res} = mg*cos(\theta) + F_{string} = ma = m(v^2/r)

And hence i isolate the force on the string

m(v^2/r) - mg*cos(\theta) = F_{string}
Tip #1: Fix your signs. The radial component of the weight opposes the tension in the string.
Someone also tried to hint to me i should use the preservation of energy, but i'm not seeing where that should enter the picture.
Tip #2: Use conservation of energy to eliminate v^2. Hint: find an expression for the KE in terms of cos\theta_0 and cos\theta.

Tip #3: Once you've eliminated v^2, replace all cos terms with the first two terms in its Taylor expansion.

[Tide]

Hint: You can find \frac {d \theta}{dt} by integrating the equation of motion
\frac {d^2 \theta}{dt^2} = - \frac {g}{L} \sin \theta
to find
\frac {d \theta}{dt} = \sqrt {\frac {2g}{L} (\cos \theta - \cos \theta_0)}
L \times d\theta / dt is the tangential velocity you need to calculate the centripetal force. You should end up with
T = 3 mg \cos \theta - 2mg \cos \theta_0
which is exact, by the way, and you can now do a taylor expansion to get the approximation!

[BLaH!]

The concepts in this problem aren't too bad but the math may be a little intimidating. What you are trying to find in this problem is T = T(\theta), the tension in the string as a function of the angle. You already correctly wrote down what this is:

T(\theta) = m\frac{v(\theta)^2}{l} - mg*cos\theta

where v(\theta) is the tangential velocity of the mass as a function of angle. At this point you don't know what v(\theta) is. This is where conservation of energy is handy. Since only conservative forces (i.e. gravity) do work on the mass, energy is conserved. I will pick U = 0 where the string is connected to the ceiling. I will take

E_{initial} = -mglcos\theta_{0}

where the mass is released from rest and

E_{final} = -mglcos\theta + \frac{1}{2}mv(\theta)^2

Equating these two energies allows us to solve for v(\theta). Thus,

v(\theta) = \sqrt{\frac{2g}{l}(cos\theta - cos\theta_{0})}

Plugging this back into the equation for the string tension gives

T = mg(3cos\theta - 2cos\theta_{0})

At this point use the small angle approxmation where

cos\theta \approx 1 - \frac{1}{2}\theta^2

That should get you the right answer.

[Dracovich]

Ok thanks guys, it's getting quite late here so i think i'll wait until tomorrow to look it all over and hopefully get the right answer :) Will report in with my findings to see if i got the right idea.

[Dracovich]

Ok so i just went through it and i seem to get it now.

I have

m(v^2/l) + mg*cos(\theta) = F_{string}

Ok, here i had a bit of a problem, i couldn't find a sensible way to represent h in E=mgh, i'd imagine that it was the height from which the bob is raised from it's lowest position right? So i thought it'd be h=l-(cos(\theta)*l), but that just gives me a lot of confusion :) I saw here above that you used mglcos(theta) to do it, and if i use that i can get the right answer. But i'd imagine that mglcos(theta) would give me l-h, but not h. If anyone could explain how that formula = h, i'd be golden :) Anyway, here's the rest of my calculations assuming that lcos(theta) = h

E(\theta_0) = -mglcos(\theta_0)

Where the angle is at the extreme of it's motion where v=0, and

E(\theta) = -mglcos(\theta) + 1/2*mv^2

At some other arbitrary point in the range of motion. Now the next step, i'm not quite sure you you guys managed to get 2g/l, but what i did to get what i wanted was:

E(\theta) = -mglcos(\theta) + 1/2*mv^2 = -mglcos(\theta_0)
mglcos(\theta)-mglcos(theta_0) = 1/2*mv^2
2glcos(\theta)-2glcos(\theta_0) = v^2
2g(cos(\theta) - cos(\theta_0)) = v^2/l

So now i have a formula for v^2/l, so i pop that into the formula i had before and get

F_{string} = m*(2g(cos(\theta) - cos(\theta_0)) + mgcos(\theta)

2mgcos(\theta) - 2mgcos(\theta_0) + mgcos(\theta) = F_{string}
mg(3mgcos(\theta)-2mgcos(\theta_0)) = F_{string}
so now if i prop the taylor approximation for theta=0 into this i get:
mg(3 - 3 * \theta^2/2 -2 + \theta^2_0/2) = F_{string}
mg(1 + \theta^2_0 - 3/2*\theta^2) = F_{string}

phew that was a lot of texing, but assuming mglcos(\theta) is right, and i'm just not seeing it o_O, this should be good :)

[Doc Al]

m(v^2/l) + mg*cos(\theta) = F_{string}
Right.
Ok, here i had a bit of a problem, i couldn't find a sensible way to represent h in E=mgh, i'd imagine that it was the height from which the bob is raised from it's lowest position right? So i thought it'd be h=l-(cos(\theta)*l), but that just gives me a lot of confusion :)
It shouldn't. But it doesn't matter where you measure the potential energy from, as long as you do it consistently. Using the lowest point as your zero, the initial PE is mg(l - l cos\theta_0). So the KE at angle \theta is 1/2 mv^2 = mg(l - l cos\theta_0) - mg(l - l cos\theta) , which simplifies nicely.

I saw here above that you used mglcos(theta) to do it, and if i use that i can get the right answer. But i'd imagine that mglcos(theta) would give me l-h, but not h. If anyone could explain how that formula = h, i'd be golden :) Anyway, here's the rest of my calculations assuming that lcos(theta) = h

E(\theta_0) = -mglcos(\theta_0)
Nothing wrong with this. Note your use of the minus sign. You are effectively taking the pivot point as your zero of PE. No problem!

[Dracovich]

Right you are, they do go out if i measure l-cos(theta)*l. I guess i should've followed that one through, i just looked at it and it looked like i'd have a lot of variables i couldn't get out of the way. But ok it all looks good now and i think i understand it all good now. Thanks a bunch guys.