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}.


Learn more about this topic:

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Kinetic Energy of Rotation

from UExcel Physics: Study Guide & Test Prep

Chapter 7 / Lesson 7
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