Why is the torsional constant of a wire inversely related to its length?

[yokan]

I came to find out through books and actual experiment that the torsional constant of a wire in a torsional pendulum is directly inverse to its length
I find the torsional constant using the rotational version of hooke's law, t=k*theta.
I have always thought that a greater length should give a greater torque (thus a higher torsional constant) because it has a higher momentum.
I tried to search for more specific explanations as to why it is a inverse relation on the internet, but without much help. The books I got in the libraries are either too simple or too advaned for me.
Any explanations would be greatly appreciated. Thx!

 

[Doc Al]

Think of it like this. The torsional constant is defined as the amount of torque needed to get a certain angular twist: k = \tau/\theta. So assume a given wire of length L has a constant k. It requires a torque \tau to produce a twist of \theta. What if I only needed a twist of \theta/2? Would you agree that I only need half the torque? (I presume you would.)

Now consider a wire of length 2L. You can think of it as being composed of two wires of length L in series. What the torsional constant of this composite wire? If I want a net twist of \theta, realize that each half of the wire only gets a twist of \theta/2. Thus the same twist (\theta) requires only half the torque. Thus the net torsional constant of a wire of length 2L is 1/2 the constant of a wire of length L. Make sense?

Note that this is the same thing that happens with springs put in series. Say I have two springs of spring constant k. If I hook them in series, what's the spring constant of the composite double spring? Figure it out the same way as I did above and you'll find that the double spring has a spring constant of k/2.

 

[yokan]

Thx a lot for you help, I got the hang of it now