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Torsional Pendulum
By Michael Doré (P904) on Monday, February 21, 2000 - 04:34 pm:

Hi, I was wondering if anyone knows why a torsional pendulum displays (approximate) simple harmonic motion? (A torsional pendulum is a string fixed at one end with an object hanging below it. The string is then twisted.)

I had a go at showing this was true using a simplified model in which:

- the string is light
- the string obeys Hooke's law
- the angle x (through which the string is twisted at any time) is small
- the height of the object at the lower end of the torsional pendulum is constant.

The approximate equation I got was:
..
x = -w2x

where w2 = pEr6/(12L3I). I = moment of inertia of the object, E = Young's modulus of string, L = length of string and r = radius of string.

This equation was approximate because I left out terms in x3 x5 etc (which I assumed were too small to be noticeable).

So according to this the motion is approximately simple harmonic with time period:

T = 2p sqrt(12L3I/pEr6)

but is this anything close to being right?

Many thanks,

Michael

By Alexander Fletcher (P1693) on Tuesday, February 22, 2000 - 01:24 pm:

Hi,
I came across a discussion of a torsional pendulum in an old (circa 1950!) mechanics book. The pendulum used wire instead of string, and the object (not the string) was twisted, but I can't see that it would hugely affect things.

If the string has length L and radius r, and the object is rotated an angle x from its position of equilibrium about the vertical line containing the string, then the couple due to the torsion of the wire tending to restore equilibrium will be xpEr4/(2L). E= Young's modulus of the string.

Thus if the object's moment of inertia is I you get

I * d2x/dt2 = -xpEr4/(2L)

Therefore

d2x/dt2 = -xpEr4/(2LI)

Hence the torsional pendulum displays SHM and the periodic time is

2psqrt(2LI/pEr4)

I hope you can ignore the differences in set-up.

By Michael Doré (P904) on Wednesday, February 23, 2000 - 08:56 am:

Hi Alexander,

Well that answer is radically different to mine so it looks like my assumption that the length of the string does not change is false. Either that or I've made a silly mistake with the algebra. I'll see if I can get anywhere without this assumption and then write back.

Thanks for your help,

Michael

By Alexander Fletcher (P1693) on Wednesday, February 23, 2000 - 11:55 am:

Hi,
I think I made a mistake in my last message. I know that the Young's Modulus of a material is the ratio of stress to strain - when the material is under tension or compression. According to my book in the case of shearing stress, the strain is measured by the distortion produced so is a different value (the material's modulus of rigidity). Since the string in the torsional pendulum does not extend but does twist, apparently this should mean that the 'E' in the equation should be replaced by the modulus of rigidity.
As for the difference in answers, perhaps using wire and twisting the object changes things?

Thanks,

Alex

By Michael Doré (P904) on Wednesday, February 23, 2000 - 04:10 pm:

Hi there!

The fact that your book was based on wires won't matter at all - my assumptions would be the same for both. Likewise for twisting the object not the wire.

I think that it should be possible to calculate the torque simply from the Young's modulus, by dividing the wire up into vertical strips and seeing how they stretch when the wire is twisted. This is what I have done. But I assumed that the length of the string doesn't change - this is probably not true. And I have just realised that even a small increase in length causes a massive error in my calculation so this must be where my argument is going wrong.

If I find a way of working it out without that assumption then I'll write back.

Thanks,

Michael

By Michael Doré (P904) on Thursday, February 24, 2000 - 11:40 am:

Oh dear! I am now making the period approximately

2 p sqrt(2LI/r2mg)

This implies that the period is approximately simply dependent of the dimensions of the system and nothing else. (Not the material of the wire or the density of the object beneath). I find this quite unlikely so will check the algebra again.

Thanks,

Michael

By Michael Doré (P904) on Saturday, February 26, 2000 - 10:15 pm:

Hi!

Right as something is clearly going wrong in my argument somewhere I'm going to write out my reasoning. If anybody can spot anything that looks even slightly dubious then I'd be delighted to hear about it.

OK, we're considering what happens when a cylindrical string/wire is twisted through an angle t. The string has length L, radius r and Young's modulus E. Let us also suppose it is extended from its natural length by a distance e. In this message I will attempt to calculate its elastic potential energy and will use this result in subsequent messages.

Consider a small prism (area A, distance p from the centre of the wire) through the wire. The key property of this prism is that its area is so small that you can neglect variations in its distance from the centre. No error will result from assuming that this is always p.

I will assume that the wire is stretched out horizontally. When the wire is twisted and stretched the parametric equation (in w) for the twisted prism is:

x = (L+e)w/t
y = p cos w
z = p sin w

In other words the prism twists into a spiral. (I assume this is a sprial anyway.) This is straightforward enough to see/prove.

Now we imagine that this spiral is made of a series of small springs. In terms of our parameter w the potential energy of such a spring should be:

½EA × extension2/natural length

Now the actual length of the spring is

sqrt(dx2 + dy2 + dz2)

by Pythag.

Now

dx = (L+e)/t dw
dy = -p sin w dw
dz = p cos w d w

Therefore the actual length of the spring is:

sqrt((L+e)2/t2+p2) dw

The natural length of the spring is the length when e = t = 0 which is:

Ldw/t

(you can convince yourself of this by thinking about the proportion of the prism we are considering).

Substitute that lot into the energy equation, integrate with respect to w between 0 and t (it is not a very hard integral at all); the elastic potential energy of a small prism when twisted through t should be:

EAt2/2L × [2L2/t2+p2+2Le/t2+e 2/t2-2L2/t2sqrt(1+2e/L + e2/L2+p2t2/L2)

I do hope I haven't made a mistake here. I've checked through quite a few times, but normally if I miss an error the first time I don't spot it on subsequent checks.

Now imagine a series of small prisms marking out a circle radius p (thickness dp). The energy is linear on A so is clearly going to be additive. Therefore the formula can be applied to a circle, radius p thickness dp. By dividing the cylinder into many such thin circles we can find the entire elastic potential energy as

ò0 r E2ppt2/2L × [2L2/t2+p2+2Le/t2+e 2/t2-2L2/t2sqrt(1+2e/L + e2/L2+p2t2/L2) dp

So far so good?

Thanks,

Michael

By Alexander Fletcher (P1693) on Sunday, February 27, 2000 - 12:16 am:

Hi,

I know you want to use Young's modulus but just as an alternative I'll write what my book did to show that a torsional pendulum executes SHM.

Right, first off you introduce a material's coefficient of rigidity. Suppose a block ABCD, fixed at BC, is acted on by a tangential or shearing force P acting along AD and producing a distortion of angle f. It's easiest to picture when ABCD is a square, say. Then the shearing strain produced is measured by tan f. If C is the modulus of shearing stress and f is the intensity of shearing stress produced, then C=f/tan f. However, since f is going to be small, f=Cf.

I know it would be nicer to keep using Young's modulus, and you probably can, but this is just the book's approach!

Next consider a bar of length l, radius R subjected to torsion by a couple applied at one end, the other end being fixed. Take a thin rectangular prism of the material running parallel to the axis of the cylinder, one edge of which is AM, A being fixed. The edge AM is now turned into the position AM'. Let the angle MAM' = df and let angle MOM' = dq, where O is the centre of the end section, and let OM = OM' = r.
Then the face of the square at MM' will be subjected to a shearing stress of intensity f; where f = cdf.

But MM' = ldf = rdq so df=rdq/l

and f = C.rdq/l ... (1)

Every portion of the area at M is twisted through the same angle, i.e. dq is constant, therefore Cdq/l is constant and therefore the shear stress is proportional to r. Now consider an annulus at distance r from O. Since the shear stress is proportional to r, it is therefore constant round the annulus, and

Shear force acting round annulus = (Cdqr/l).2prdr

Moment of this force about the axis of the cylinder = (Cdqr/l).2prdr.r
= (2pCdq/l).r3 dr

Hence, total moment of resistance of the shaft to torsion

= (2pCdq/l)ò0 R r3 dr

= (pC/l) . ½R4 . dq

From this we know that in the case of the torsional pendulum (of length l, radius r), the couple due to the torsion of the wire tending to restore quilibrium will be (Cpr4/2l)q. Following my first message (except the 'E' which should have been a 'C') the periodic time of the pendulum is 2p sqrt(2lI/Cpr4).

That's the book's method of showing SHM anyway, just in case you're interested.

Thanks,

Alex

By Michael Doré (P904) on Sunday, February 27, 2000 - 05:57 pm:

Hi Alexander, thanks for that.

Yes I find it rather worrying that no textbooks seem to think that the period can be worked out from the Young's modulus (along with the dimensions of the system, and the mass and MI of the object beneath). I'm probably going about this completely the wrong way, though my argument does seem plausible.

I don't think that the book is showing that the motion is SHM from first principles as you have to assume the equation:

f = C tan(f)

i.e. f is proportional to tan(f). I'm not sure why this should be true.

I'll post the next part of my argument shortly.

Thanks,

Michael

P.S. Can you see any problem with my reasoning above thus far? Re-reading it, it doesn't seem very clear, so I'd welcome advice on how to expand it.