A uniform wheel of mass 10.0 kgand radius 0.400 mis mounted rigidly on a mass less axle through its center (Fig. 11-62). The radius of the axle is 0.200 m, and the rotational inertia of the wheel–axle combination about its central axis is0.600 kg.m². The wheel is initially at rest at the top of a surface that is inclined at angle$\mathbf{\theta }\mathbf{=}\mathbf{30}\mathbf{.}{\mathbf{0}}^{\mathbf{°}}$. with the horizontal; the axle rests on the surface while the wheel extends into a groove in the surface without touching the surface. Once released, the axle rolls down along the surface smoothly and without slipping. When the wheel–axle combination has moved down the surface by 2.00 m, what are (a) its rotational kinetic energy and (b) its translational kinetic energy?

A wheel of mass M=10 kg and radius is R=0.400 m.

The radius of theaxleis R=0.200 m.

The rotational inertia ofwheel-axlecombination about its central axis 0.600 kg.m².

Top of the surface inclined at $\mathrm{\theta }=30.0^{°}$.

Using the formula $\mathbf{∆}\mathbf{U}\mathbf{=}\mathbf{∆}\mathbf{K}$,find translational and rotational kinetic energy of the wheel-axle combination when it is moved down the surface by 2.00 m.

Formulae are as follow:

$$\begin{gathered} \mathrm{U}=\mathrm{mgh}=\mathrm{mgd}\mathrm{sin\theta } \\ \mathrm{K}=\frac{1}{2}{\mathrm{mv}}^2 \\ \mathrm{K}=\frac{1}{2}{\mathrm{I\omega }}^2 \end{gathered}$$

where,$\mathrm{\omega }$is angular frequency, L is angular momentum, Iis moment of inertia, m is mass, v is velocity, gis an acceleration due to gravity, h is height, d is distance, U is potential energy and K is kinetic energy.

The wheel rolls down along the incline by distance d, the height is $\mathrm{dsin\theta }$.

So, potential energy lost is$∆\mathrm{U}=\mathrm{mgd}\mathrm{sin\theta }$.

According to energy conservation law, this potential energy would be converted to kinetic energy,

$\begin{array}{rcl}-∆\mathrm{U}& =& ∆\mathrm{K}=∆{\mathrm{K}}_{\mathrm{trans}}+∆{\mathrm{K}}_{\mathrm{rot}}\\ \mathrm{mgd}\mathrm{sin\theta }& =& \frac{1}{2}{\mathrm{mv}}^2+\frac{1}{2}{\mathrm{I\omega }}^2\\ \mathrm{mgd}\mathrm{sin\theta }& =& \frac{1}{2}\mathrm{m}{\left(\mathrm{r\omega }\right)}^2+\frac{1}{2}{\mathrm{I\omega }}^2\\ \mathrm{\omega }& =& \frac{\mathrm{v}}{\mathrm{r}}\\ \mathrm{mgd}\mathrm{sin\theta }& =& \frac{1}{2}{\mathrm{I\omega }}^2\left(\frac{{\mathrm{mr}}^2}{\mathrm{I}}+1\right)\\ \mathrm{mgd}\mathrm{sin\theta }& =& ∆{\mathrm{K}}_{\mathrm{rot}}\left(\frac{{\mathrm{mr}}^2}{\mathrm{I}}+1\right)\\ ∆{\mathrm{K}}_{\mathrm{rot}}& =& \frac{\mathrm{mgd}\mathrm{sin\theta }}{\left({\displaystyle \frac{{\mathrm{mr}}^2}{\mathrm{I}}}+1\right)}\\ & =& \frac{\left(10.0\mathrm{kg}\right)\times \left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(2.00\mathrm{m}\right)\sin \left(30.0^{°}\right)}{\left[{\displaystyle \frac{\left(10.0\mathrm{kg}\right){\left(0.200\mathrm{m}\right)}^2}{0.600\mathrm{kg}.{\mathrm{m}}^2}}+1\right]}\\ ∆{\mathrm{K}}_{\mathrm{rot}}& =& 58.8\mathrm{J}\end{array}$.

Hence,rotational kinetic energy of the wheel-axle combination when it moved down the surface by 2.00 m is $∆{\mathrm{K}}_{\mathrm{rot}}=58.8\mathrm{J}$.

Now,

$\begin{array}{rcl}∆{\mathrm{K}}_{\mathrm{trans}}& =& ∆\mathrm{K}-∆{\mathrm{K}}_{\mathrm{rot}}\\ ∆\mathrm{K}& =& ∆\mathrm{U}\\ & =& \mathrm{mgd}\sin 30.0^{°}\\ & =& \left(10\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(2\mathrm{m}\right)\left(\sin {30}^{°}\right)\\ & =& 98\mathrm{J}\\ ∆{\mathrm{K}}_{\mathrm{trans}}& =& 98\mathrm{J}-58.8\mathrm{J}\\ ∆{\mathrm{K}}_{\mathrm{trans}}& =& 39.2\mathrm{J}\\ & & \end{array}$

Hence, translational kinetic energy of the wheel-axle combination when it moved down the surface by 2.00 m is$∆{\mathrm{K}}_{\mathrm{trans}}=39.2\mathrm{J}$

Therefore, using the law of conservation of energy, the rotational and kinetic energy of the object when it is rolling down the incline can be found.