Recent content by BruceSpringste

Homework Statement We have experimented on rolling cylinders and have gotten the function for a rolling cylinder but have gotten a strange constant from experiment. Homework Equations t = 2\sqrt{\frac{l}{g}(1+\frac{r}{R})\frac{1}{\sin{\theta}}}-0.5 The Attempt at a Solution Is this constant...

Yes I understand that. But how come you didn't write D/d instead? I thought maybe you had some physical reasoning behind the choice of d/D instead of D/d.

Thank you, I think I understand your line of thought now! Edit: Altough the fact that you wrote d/D bothers me a bit. It appealed to you aestheticly. Imagine I could have used physics when I were done with the analysis. Could I have come to the conclusion d/D?

You're not using Buckinghams I am guessing? Also the inner diameter (d) does change the moment of inertia?

I kind of got to the same conclusion. However I had (\frac{d}{l}, v, \frac{D}{l})How did you come to that conclusion? Also how do you scale with l? Shouldnt time be scaled with t\sqrt{\frac{g}{sin(v)*l}}. Since g acts vertically I mean.

Hello! I am doing an experiment on rolling cylinders with both an inner diameter and outer diameter. I.E. they are not solid. I have to determine the time it takes for a cylinder to roll down an inclined slope. I need to do a dimensional analysis with Buckinghams PI-Theorem but I am stuck and...

Sorry for not replying. I finished the experiment with a correct analysis and totally forgot this post. To anyone viewing this in the future, use Buckinghams PI - Theorem. From that you will reduce the amount of parameters you need to study in order to come to find the formula. The point of the...

First and foremost thank you for taking your time and helping me. 1. No function is given. We simply made a dimensional analysis of the physical parameters that are involved, mass, time and length. And tried to make sense of it dimensionally. The form of this function does not fit the data...

If \begin{equation} T=A\times{L^a}\times{M^b}\times{(\frac{L}{T^2})^c}\times{(\frac{ML}{T^2})^d} \end{equation} Then 1=-2d-2c 0=a+c+d 0=b+d which has infinite solutions. Edit: However if we could assume that the value of a and c are unchanged from the analysis of an ideal pendulum it is...

B/L is the term which equals that of an ideal pendulum whilst the added term C*L should be that of the moment of the inertia?

Exactly. The form of the equation is consistent with the ideal pendulum. But the pendulum we are studying has a spread mass and not a point mass. When we linearised our data we get an extra constant which we haven't taken into account. With the help of a proper dimensional analysis we can find...

Yes that is what I am calling inertia. And it seems that this was our problem. If we take into account the dimensions of inertia our intial analysis will then make sense. Or am I mistaken? The dimensions of inertia are [M][T^(-2)] which we haven't taken into account.

\begin{equation} m\times{\overrightarrow{a}} \end{equation}

The inertia is the resistance of any physical object to any change in its state of motion. With the data from the laboration we get an equation with MATLAB by linearisation so we made the change to the dimensional analysis to get: \begin{equation} T = \sqrt{\frac{B}{L} + CL} \end{equation}...

The inertia of the pendulum. But how do I include this? As a factor?