The impact of mass distribution on time taken for a cylinder to roll down a ramp

Question

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.

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:

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

so

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

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.