Question

When a solid sphere is rolling along a level surface the percentage of its total kinetic energy that is transitional is:
A) 29%
B) 71%
C) 60%
D) 40%

Answer 1

Hint: Here a solid sphere is rolling along a straight surface, hence it will have kinetic energy. Now the kinetic energy of the sphere will have two components, rotational and transitional. And we have to find the percentage of transitional components in total kinetic energy. By using the formula of rotational and translational kinetic energy we can solve the given question.
Formula used:
\[\begin{align}
  & K.E{{.}_{trans}}=\dfrac{1}{2}m{{v}^{2}} \\
 & K.E{{.}_{rotat}}=\dfrac{1}{2}I{{\omega }^{2}} \\
 & I=\dfrac{2}{5}m{{r}^{2}} \\
 & \omega =\dfrac{v}{r} \\
\end{align}\]

Complete answer:
Let us first draw a simple diagram for the given question.

Here a solid sphere is rolling over a plane. Let's say the sphere has the radius r and as it is moving forward it has velocity which will be always tangential to the surface of the plane whereas due to its rotational motion it will have angular velocity ω. Similarly, it will have two components of the kinetic energy, one will be due to its translational motion and other due to its rotational motion. And the total energy of the sphere will be the sum of rotational kinetic energy and translational kinetic energy.
Hence total kinetic energy of the sphere will be given as
\[K.E{{.}_{total}}=K.E{{.}_{trans}}+K.E{{.}_{rotat}}\]
Now formula of transitional energy is given as
\[K.E{{.}_{trans}}=\dfrac{1}{2}m{{v}^{2}}\]
Where m is the mass and v is velocity.
And the formula of rotational energy is given as
\[K.E{{.}_{rotat}}=\dfrac{1}{2}I{{\omega }^{2}}\]
Where I is the moment of inertia and ω is angular velocity. Hence total kinetic energy will be given as
\[\begin{align}
  & K.E{{.}_{total}}=\dfrac{1}{2}m{{v}^{2}}+\dfrac{1}{2}I{{\omega }^{2}} \\
 & \Rightarrow K.E{{.}_{total}}=\dfrac{1}{2}\left( m{{v}^{2}}+I{{\omega }^{2}} \right) \\
\end{align}\]
Now moment of inertia for the solid sphere is given as
\[I=\dfrac{2}{5}m{{r}^{2}}\]
Substituting this value of moment of inertia in above equation, we get
\[K.E{{.}_{total}}=\dfrac{1}{2}\left( m{{v}^{2}}+\dfrac{2}{5}m{{r}^{2}}{{\omega }^{2}} \right)\]

We know that angular velocity of a body is given as
\[\begin{align}
  & \omega =\dfrac{v}{r} \\
 & \Rightarrow v=\omega r \\
\end{align}\]
Substituting value of ωr as v in the above equation we get
\[\begin{align}
  & K.E{{.}_{total}}=\dfrac{1}{2}\left( m{{v}^{2}}+\dfrac{2}{5}m{{v}^{2}} \right) \\
 & \Rightarrow K.E{{.}_{total}}=\dfrac{1}{2}m{{v}^{2}}\left( 1+\dfrac{2}{5} \right) \\
 & \Rightarrow K.E{{.}_{total}}=\dfrac{1}{2}m{{v}^{2}}\left( \dfrac{7}{5} \right) \\
 & \Rightarrow K.E{{.}_{total}}=\dfrac{7}{10}m{{v}^{2}} \\
\end{align}\]
Now the contribution of translational kinetic energy in total kinetic energy can be found dividing the translational kinetic energy by total kinetic energy.
\[ \dfrac{K.E{{.}_{trans}}}{K.E{{.}_{total}}}=\dfrac{\dfrac{1}{2}m{{v}^{2}}}{\dfrac{7}{10}m{{v}^{2}}} \]
\[ \Rightarrow \dfrac{K.E{{.}_{trans}}}{K.E{{.}_{total}}}=\dfrac{1}{2}\times \dfrac{10}{7} \]
\[ \Rightarrow \dfrac{K.E{{.}_{trans}}}{K.E{{.}_{total}}}=\dfrac{10}{14} \]
\[ \Rightarrow \dfrac{K.E{{.}_{trans}}}{K.E{{.}_{total}}}=0.71 \]
 \[ \Rightarrow \dfrac{K.E{{.}_{trans}}}{K.E{{.}_{total}}}=71% \]

So, the correct answer is “Option B”.

Note:
We can see that the contribution of translational kinetic energy of the system will depend on the shape or type of object. As for different shapes, moments of inertia will be different and so the value of rotational kinetic energy will change resulting in a change of total kinetic energy. Although in any case most of the kinetic energy is due to translational motion.