Measuring Torque: Ideas & Solutions for Lab Projects

[MMS]

Hello everyone.
I'm working on a physics lab project and I've reached a barrier.
I need to measure the torque on a closed loop in a magetic field and I can't reach a way to do so. Of course, excluding all the torque measuring devices and such which I believe we don't have at the lab.

Any ideas on how to measure it?Thank you.

 

[ShayanJ]

I think you can use a spring with a known spring constant(which should be chosen in accordance with the range of torque being applied to the loop, you should have at least a very rough estimate of the amount of torque). Then you attach the loop to one end of the spring and fix the other end. Then you measure the elongation of the spring. Now you calculate \frac 1 2 k \delta x^2 which should be approximately equal to \int_{\theta_i}^{\theta_f} \tau d\theta. If the rotation angle is small, I think you can take the torque to be constant and that gives you the amount of torque.

 

[MMS]

Interesting.
I can estimate what the torque should be by calculations so I believe that shouldn't be a problem. Just a question: how can I deal with the integral on the torque that you've given with large angles? Why can't I treat tge torque as a constant in both cases (small and large)?

 

[ShayanJ]

Take the interval (a,b) where b=a+H. Now I want to calculate the integral \int_a^b f(x) dx(f being any function). Now I substitute x=a+\frac H 2 +h \Rightarrow dx=dh.
<br /> \int_{-\frac H 2}^{\frac H 2} f(a+\frac H 2+h) dh=\int_{-\frac H 2}^{\frac H 2} [\sum_{n=0}^\infty \frac{ f^{(n)}(a+\frac H 2)}{n!} h^n] dh=\int_{-\frac H 2}^{\frac H 2} [f(a+\frac H 2)+O(h)] dh=f(a+\frac H 2)H+O(H^3) \\ \Rightarrow f(a+\frac H 2)=\frac{1}{H} \int_a^b f(x)dx+O(H^2)<br />
So by taking \frac{1}{\theta_f-\theta_i}\int_{\theta_i}^{\theta_f} \tau d\theta, to be the torque at \theta=\frac{\theta_f+\theta_i}{2}, you have an error of the order (\theta_f-\theta_i)^2.
All of this is because the torque may not be constant during the rotation and how it changes is unknown to us. You may think its actually constant from some calculations but that only makes the approximation better because in reality, things aren't as clean as on the paper!

 

[MMS]

Thanks. I'll make sure to go over it as when I get home (I'm on my phone).
Just a couple more questions if I may.
1. Did you mean to connect the spring directly to the loop or by a string or something?
2. Aren't there other things that contribute to the work in the system (except for the spring)?

 

[ShayanJ]

1. Did you mean to connect the spring directly to the loop or by a string or something?

Connect them directly! Otherwise those things in between, will get some energy and that worsens the approximation.

2. Aren't there other things that contribute to the work in the system (except for the spring)?

I think the spring's potential energy is, to a good approximation, the only thing that gets the loop's rotational energy.