Generalize Problem $\mathbf{14}$to any mass $\mathbf{M}$of circular cross-section and moment of inertia $\mathbf{I}$. Consider a hoop, a disk, a spherical shell, a solid spherical ball; order them as to which would first reach the bottom of the inclined plane. (For moments of inertia, see Chapter $\mathbf{5}$, Section $\mathbf{4}$.)

An inclined plane is also termed as a ramp. It is a smooth supporting surface that is inclined at an angle, including one end that is beyond another, and is used to actually increase or reduce weight. Heavy weights are moved over vertical obstacles using inclined planes.

Given that the mass is$\mathrm{M}$ of circular cross-section and the moment of inertia is $\mathrm{I}$.

As the hoop is translating and rotating.

Thus, the kinetic energy will be the sum of the translational and rotational part.

This means that,

$$\begin{gathered} \mathrm{T}={\mathrm{T}}_{\mathrm{trans}}+{\mathrm{T}}_{\mathrm{rot}} \\ =\frac{1}{2}{\mathrm{Mv}}^2+\frac{1}{2}{\mathrm{I\omega }}^2 \\ =\frac{1}{2}\mathrm{M}{\dot{\mathrm{y}}}^2+\frac{1}{2}\mathrm{I}{\dot{\mathrm{\theta }}}^2 \end{gathered}$$

Since the translational motion is inclined along the $\mathrm{y}$axis

And as there is no slipping, thus

$$\begin{gathered} \mathrm{y}=\mathrm{a\theta }+\mathrm{const}. \\ \stackrel{.}{\mathrm{y}}=\mathrm{a}\stackrel{.}{\mathrm{\theta }} \end{gathered}$$

Since the above constraint expresses the Lagrangian in the terms of $\mathrm{y}$of $\mathrm{\theta }$.

Then the kinetic energy will be:

$$\begin{gathered} \mathrm{T}=\frac{1}{2}\left(\mathrm{M}{\dot{\mathrm{y}}}^2+\mathrm{I}{\dot{\mathrm{\theta }}}^2\right) \\ =\frac{1}{2}\left(\mathrm{M}{\dot{\mathrm{y}}}^2+\mathrm{I}\frac{{\dot{\mathrm{y}}}^2}{{\mathrm{a}}^2}\right) \end{gathered}$$

Now due to the gravity and the elevation of the centre of the mass, the potential energy of the hoop $\mathrm{z}$is related to the position $\mathrm{y}$, given by:

$\mathrm{z}=\mathrm{ysin\alpha }$

It is a known fact that the latter follows from the geometry of the inclined plane and also Lagrangian is defined as $\mathrm{L}=\mathrm{T}-\mathrm{V}$, thus Lagrangian will be given by:

$\mathrm{L}=\mathrm{M}{\dot{\mathrm{y}}}^2-\mathrm{Mgysin\alpha }$

So, the Lagrangian will be given by:

$\mathrm{L}=\mathrm{M}{\dot{\mathrm{y}}}^2-\mathrm{Mgysin\alpha }$

Now the Euler equation reads:

$\frac{\mathrm{d}}{\mathrm{dt}}\frac{\partial \mathrm{L}}{\partial \dot{\mathrm{y}}}-\frac{\partial \mathrm{L}}{\partial \mathrm{y}}=0$

Then, the required derivatives will be:

Then the equation of motion from the Euler equation and the above equations will be:

So, Lagrange’s equation is $2\mathrm{M}\ddot{\mathrm{y}}+\mathrm{Mgsin\alpha }=0$.

As the acceleration determines velocity and the velocity determines the mass which will reach the bottom first and the acceleration$\ddot{\mathrm{y}}$from the previous equation.

This means that,

Since the moment of inertia is the smallest, it concludes the largest absolute value of acceleration and makes sense since the inertia is a measure of the body’s resistance to acceleration.

Then the object that reaches the bottom first will be:

So, the Spherical shell will arrive first at the bottom of the inclined planed and then followed by the disk and then the hoop