I'm getting some weird results from a calculation I'm doing and quite honestly, I'm pretty sure it's due to human error. I do have an apparatus involved for the experimental process for my lab but I don't think it's what's causing the problem. I've come to the conclusion that my notes do not contain the right equations for these calculations.

For the theoretical masses, I know the total mass of the point masses and the distance from the axis to the masses. I'm pretty sure the equation of this is either 1/2mr^2 or just mr^2. The 1/2Mr^2 is from my notes but I think the correct answer would be mr^2. However the hanging mass lies on an apparatus that is a cylinder so 1/2Mr^2 could be correct.

The experimental part of the lab involved an apparatus that looked like a wheel which lowered a hanging mass on a string by turning the wheel (also by gravity). For the point mass and apparatus combined, I know the hanging mass, slope, and radius. I also know this data for the apparatus. I started out by finding the Force (which I was told is also equal to the torque) using the equation F=m(g-a). I know a is acceleration, but I'm not exactly sure how that factors out. From there I'd probably use τ = I α but I know the angular acceleration nor how to calculate it.

The next part of my lab involves calculating the experimental rotational inertia of the ring and disk, and I know the hanging mass, slop and radius of each one. Would that be where 1/2Mr^2 comes in?

  • $\begingroup$ Do you mean something like this, because you mention a slope? Since your description of the experimental setup it not 100% clear to me. $\endgroup$
    – fibonatic
    Jan 7, 2014 at 17:28
  • $\begingroup$ Sort of, except exactly downward. $\endgroup$
    – Genevieve
    Jan 7, 2014 at 18:50
  • $\begingroup$ But what do you mean with the slope, or do you mean that you know that the slope it vertical? $\endgroup$
    – fibonatic
    Jan 7, 2014 at 18:55
  • $\begingroup$ To be honest, I'm not completely sure. We used some sort of a program to measure it. $\endgroup$
    – Genevieve
    Jan 7, 2014 at 20:00
  • $\begingroup$ It was similar to the device mentioned in this pdf: physics.gmu.edu/phys161/spring2009/… $\endgroup$
    – Genevieve
    Jan 7, 2014 at 20:08

1 Answer 1


Moment of inertia
The definition of (mass) moment of inertia of a point mass is $$ I=r^2m $$ However in the real world you don't encounter point masses, but objects with non-zero volume (finite density). And leads to an integral to determine moment of inertia $$ I=\int_m{r^2dm}=\int_V{\rho(r)r^2dV}=\int_x{\int_y{\int_z{\rho(x,y,z)(x^2+y^2+z^2)dz}dy}dx} $$ The solutions of this integral of a few bodies, with constant non-zero density within geometric volume and zero density outside of it, can be found here. For example the moment of inertia of thin rod rotating around its center of mass is equal to $I=\frac{mL^2}{12}$ and for a solid cylinder $I=\frac{mL^2}{2}$.
Experimental setup
In your experimental a string is on one end connected to the hanging mass, lead over the pulley and then wounded around a drum (the other end is also connected to the drum). This drum is of the object from which you would like to determine its moment of inertia and it is assumed that it can rotate freely (without slip) around its axis.
According to your documentation you measure how far the pulley has rotated, I will call this angle $\theta$, and its first and second time derivative $\omega=\dot{\theta}$ and $\alpha=\ddot{\theta}$.
The displacement of the hanging mass is related to the angular displacement of the pulley and its radius, $r_p$, assuming that the string does not slip, so $$ s=r_p\theta $$ where $s$ is the vertical downward displacement of the hanging mass.
This displacement is equal to the amount of string unrolled from the drum (assuming that the string is not elastic), which means that the angular displacement of the object from which you would like to determine the moment of inertia, I will call this $\theta_I$, can be calculated from this the other way around using the radius of the drum $r_d$ $$ s=r_p\theta=r_d\theta_I\rightarrow\theta_I=\frac{s}{r_d}=\frac{r_p}{r_d}\theta $$ This linear correlation also applies to the $\omega$ and $\alpha$.
The only force applied on this system (which can perform work) is gravity on the hanging mass. Using all this you can derive the equation of motion (possibly using free body diagrams and tension in the string).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.