A cylinder ( I = 1/2 MR^2), a sphere ( I = 2/5 MR^2), and a hoop ( I = MR^2) roll without...

Question:

A cylinder ( I = 1/2 MR{eq}^2 {/eq}), a sphere ( I = 2/5 MR{eq}^2 {/eq}), and a hoop ( I = MR{eq}^2 {/eq}) roll without slipping down the same incline, beginning from rest at the same height. All three objects share the same radius and total mass. Which object has the greatest translational kinetic energy at the bottom of the incline?

Rolling Body Kinetic Energy:

If a solid or rigid body (a sphere, a cylinder, a hoop or a disc) of mass {eq}M {/eq} and of radius {eq}R {/eq} rolls without slipping with linear speed {eq}V {/eq}.

Then, it will have total kinetic energy given by:

{eq}\displaystyle{KE = \frac{1}{2}MV^{2}\left(1 + \frac{k^2}{R^2}\right)} {/eq}.

({eq}k {/eq} is radius of gyration of the body)

Transnational kinetic energy of the body is given by:

{eq}\displaystyle{KE_T = \frac{1}{2}MV^2} {/eq}.

Therefore, total kinetic energy can be expressed as:

{eq}\displaystyle{KE = KE_T\left(1 + \frac{k^2}{R^2}\right)} {/eq}.

Therefore, transnational kinetic energy can be expressed as:

{eq}\displaystyle{KE_T = \frac{KE}{1 + \frac{k^2}{R^2}}} {/eq}.

Let radius of gyration for the cylinder, the sphere and the hoop be {eq}\color{blue}{k_c,\, k_s} {/eq} and {eq}\color{blue}{k_h} {/eq} respectively.

For the cylinder: {eq}\color{blue}{\frac{k^2_c}{R^2} = \frac{1}{2}} {/eq}

For the sphere: {eq}\color{blue}{\frac{k^2_s}{R^2} = \frac{2}{5}} {/eq}.

For the hoop: {eq}\color{blue}{\frac{k^2_h}{R^2} = 1} {/eq}.

The cylinder, the sphere and the hoop start rolling from the same height from rest, having equal mechanical energy as they have equal mass and their total kinetic energy at the bottom will be equal total kinetic energy {eq}\color{blue}{K} {/eq} (say).

Then,

Transnational kinetic energy of the cylinder is given by:

{eq}\displaystyle{\begin{align*} \color{blue}{T_c} &= \color{blue}{\frac{K}{1 + \frac{k^2_c}{R^2}}}\\ &= \color{blue}{\frac{K}{1 + \frac{1}{2}}}\\ &= \color{blue}{\frac{2}{3}K}\\ \end{align*}} {/eq}

Transnational kinetic energy of the sphere is given by:

{eq}\displaystyle{\begin{align*} \color{blue}{T_s} &= \color{blue}{\frac{K}{1 + \frac{k^2_s}{R^2}}}\\ &= \color{blue}{\frac{K}{1 + \frac{2}{5}}}\\ &= \color{blue}{\frac{5}{7}K}\\ \end{align*}} {/eq}

Transnational kinetic energy of the hoop is given by:

{eq}\displaystyle{\begin{align*} \color{blue}{T_h} &= \color{blue}{\frac{K}{1 + \frac{k^2_h}{R^2}}}\\ &= \color{blue}{\frac{K}{1 + 1}}\\ &= \color{blue}{\frac{1}{2}K}\\ \end{align*}} {/eq}

Now, we see:

{eq}\displaystyle{\color{red}{\frac{5}{7}K\, >\, \frac{2}{3}K\, >\, \frac{1}{2}K} \implies \color{red}{T_s\, >\, T_c\, >\, T_h}} {/eq}

Hence, the sphere will have the greatest transnational kinetic energy at the bottom of the incline.