# An 8.10-cm-diameter, 340 g solid sphere is released from rest at the top of a 1.80-m-long,...

## Question:

An 8.10-cm-diameter, 340 g solid sphere is released from rest at the top of a 1.80-m-long, 18.0-degree incline. It rolls, without slipping, to the bottom. What is the sphere's angular velocity at the bottom of the incline? What fraction of its kinetic energy is rotational?

## Rotational kinetic energy:

Rotational kinetic energy is a type of kinetic energy. The rotational kinetic energy depends upon the moment of the inertia and the angular speed of the object. The expression for the rotational kinetic energy is,

{eq}K = \dfrac{1}{2}I{\omega ^2} {/eq}

Here, the moment of inertia is {eq}I {/eq} and the angular velocity is {eq}\omega {/eq}.

## Answer and Explanation:

**Given Data**

- The diameter of the sphere is: {eq}d = 8.10\;{\rm{cm}} {/eq}.

- The mass of the sphere is: {eq}m = 340\;{\rm{g}} {/eq}.

- The height is: {eq}{h_0} = 1.80\;{\rm{m}} {/eq}.

- The angle is: {eq}\theta = 18^\circ {/eq}.

The expression for the vertical height is,

{eq}h = {h_0}\sin \theta {/eq}

Substitute the given values.

{eq}\begin{align*} h &= 1.80\sin 18^\circ \\ &= 0.5562\;{\rm{m}} \end{align*} {/eq}

The expression for the energy conservation is,

{eq}\begin{align*} \Delta P &= {K_1} + {K_2}\\ mgh &= \dfrac{1}{2}m{v^2} + \dfrac{1}{2}I{\omega ^2}\\ mgh &= \dfrac{1}{2}m{\left( {r\omega } \right)^2} + \dfrac{1}{2}\left( {\dfrac{2}{5}m{r^2}} \right){\omega ^2}\\ gh &= 0.5{r^2}{\omega ^2} + 0.2{r^2}{\omega ^2}\\ {\omega ^2} &= \dfrac{{gh}}{{0.7{r^2}}}\\ \omega &= \sqrt {\dfrac{{gh}}{{0.7{r^2}}}} \end{align*} {/eq}

Substitute the given values.

{eq}\begin{align*} \omega &= \sqrt {\dfrac{{9.81 \times 0.5562}}{{0.7{{\left( {\dfrac{{8.10\;{\rm{cm}}}}{2}\left( {\dfrac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)} \right)}^2}}}} \\ &= \sqrt {\dfrac{{5.456322}}{{1.148175 \times {{10}^{ - 3}}}}} \\ &= 68.935\;{\rm{rad/s}} \end{align*} {/eq}

Thus, the angular velocity is {eq}68.935\;{\rm{m/s}} {/eq}.

The expression for the total energy is,

{eq}P = mgh {/eq}

Substitute the given values.

{eq}\begin{align*} P &= 340\;{\rm{g}}\left( {\dfrac{{1\;{\rm{kg}}}}{{1000\;{\rm{g}}}}} \right) \times 9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}} \times 0.5562\;{\rm{m}}\\ &= 1.855\;{\rm{J}} \end{align*} {/eq}

The expression for the rotational energy is,

{eq}E = \dfrac{1}{5}m{r^2}{\omega ^2} {/eq}

Substitute the given values.

{eq}\begin{align*} E &= \dfrac{1}{5}340\;{\rm{g}}\left( {\dfrac{{1\;{\rm{kg}}}}{{1000\;{\rm{g}}}}} \right){\left( {\dfrac{{8.10\;{\rm{cm}}}}{2}\left( {\dfrac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)} \right)^2}{\left( {68.935\;{\rm{m/s}}} \right)^2}\\ &= 0.530\;{\rm{J}} \end{align*} {/eq}

The expression for the fraction is,

{eq}F = \dfrac{E}{P} {/eq}

Substitute the given values.

{eq}\begin{align*} F &= \dfrac{{0.53}}{{1.855}}\\ &= 0.2857 \times 100\% \\ &= 28.57\% \end{align*} {/eq}

Thus, the fraction of its kinetic energy is {eq}28.57\% {/eq}.

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