I am carrying out an experiment to test out the impact of mass distribution on time taken for a cylinder to roll down a ramp. I have kept the overall mass of the cylinder constant, as well as the distance the cylinder covers.

After plotting a graph of distance against time, the graph definitely matches the theory, as it showed that the cylinder with higher moment of inertia took a longer time to roll down ramp. However, I am not able to find the relationship between both. The original graph was a slight curve, however a log-log graph displayed no exponential relationship.

When plotting time against distance squared, the graph showed a straight line. I was wondering whether I can say that moment of inertia is proportional to time, as the formula for moment of inertia is directly proportional to r squared. Any suggestions about what relationship is expected is really appreciated.

If mass stays constant, however the only thing that changes is the mass distribution in equal intervals away from the axis of rotation, what is the expected relationship.

  • $\begingroup$ Do you fix the radius of the cylinder in different attempts of your experiment? I am not sure what the distance means in your experiment, but for a given cylinder, height should be proportional to t^2. Furthermore, for a momentum of inertia I, radius of cylinder r, time t height h, you should have $I/r^2 = k t^ 2/ h$, where k is a proportionality constant depending on the plane angle, the mass of the cylinder and g. (You can check this with dimensional analysis). If you want more details, I will make a full response. $\endgroup$ Oct 11, 2021 at 15:55
  • $\begingroup$ Hello, yes the radius is fixed throughout the whole experiment, and so is the overall mass, only the mass distribution is changed. For distance, I mean the distance of mass from the axis of rotation. For this experiment, I stuck 2 cylinders next to each other, only connected by two identical masses placed at identical spacing from axis of rotation. I am trying to see what the expected relationship should be between time and distance mass was placed from axis of rotation. More detail would be very appreciated. Thank you $\endgroup$
    – Physics
    Oct 11, 2021 at 19:36
  • $\begingroup$ Check @JEB 's answer for a complete solution, mine is very similar $\endgroup$ Oct 11, 2021 at 20:09
  • $\begingroup$ Ok sure, thanks. $\endgroup$
    – Physics
    Oct 12, 2021 at 18:25
  • $\begingroup$ Hello, quick follow-up question. What exactly is effective inertial mass, and what are h coordinates. Thanks $\endgroup$
    – Physics
    Oct 13, 2021 at 16:54

1 Answer 1


So you have fixed mass, $m$, which is distributed to give a moment of inertia, $I$. If the cylinder has radius $R$, the you could say:

$$ 0\le I \le mR^2$$

however, a construction like a train-wheel, means $I$ can be unbounded.

Presumable $I$ is diagonal, and the balanced, so that the gravitational potential energy is just:

$$ U = mgh $$

where $h$ is vertical distance to the axis. If the cylinder roll at a speed $v$, then there are two kinetic energy terms. One from the linear motion:

$$ T_l = \frac 1 2 m v^2 $$

and another from rotation:

$$ T_r = \frac 1 2 I\omega^2 $$

With the constraint $v=\omega R$, the total kinetic energy is:

$$ T = \frac 1 2 m v^2 + \frac 1 2 I v^2/R^2 = \frac 1 2 v^2(m + \frac I{R^2})$$

If the angle of the ramp is $\theta$, then:

$$ \dot h = v\sin\theta $$


$$ T = \frac 1 2 \dot h ^2\frac{(m + \frac I{R^2})}{\sin^2\theta} =\frac 1 2 m_{\rm eff}\dot h^2$$

where the effective inertial mass in $h$ coordinates is:

$$m_{\rm eff} = \frac{(m + \frac I{R^2})}{\sin^2\theta}$$

Combine that with the potential energy:

$$ U = mgh = m_{\rm eff}g_{\rm eff} h $$

where now we've made and "effective" gravitational acceleration:

$$ g_{\rm eff} = g \times \frac{\sin^2\theta}{1 + \frac I{mR^2}} $$

giving standard equation in $(h, \dot h)$, in which $m_{\rm eff}$ cancel on both sides, allowing you to solve for time.


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