91Ó°ÊÓ

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 $\mathrm{θ}=30.0^{°}$.

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:

$$\begin{gathered} \mathrm{U}=\mathrm{mgh}=\mathrm{mgd}\mathrm{²õ¾\pm ²Ôθ} \\ \mathrm{K}=\frac{1}{2}{\mathrm{mv}}^2 \\ \mathrm{K}=\frac{1}{2}{\mathrm{\pm õÓ\neg }}^2 \end{gathered}$$

where,$\mathrm{Ó\neg }$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 $\mathrm{d²õ¾\pm ²Ôθ}$.

So, potential energy lost is$∆\mathrm{U}=\mathrm{mgd}\mathrm{²õ¾\pm ²Ôθ}$.

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

$\begin{array}{rcl}-∆\mathrm{U}& =& ∆\mathrm{K}=∆{\mathrm{K}}_{\mathrm{trans}}+∆{\mathrm{K}}_{\mathrm{rot}}\\ \mathrm{mgd}\mathrm{²õ¾\pm ²Ôθ}& =& \frac{1}{2}{\mathrm{mv}}^2+\frac{1}{2}{\mathrm{\pm õÓ\neg }}^2\\ \mathrm{mgd}\mathrm{²õ¾\pm ²Ôθ}& =& \frac{1}{2}\mathrm{m}{\left(\mathrm{rÓ\neg }\right)}^2+\frac{1}{2}{\mathrm{\pm õÓ\neg }}^2\\ \mathrm{Ó\neg }& =& \frac{\mathrm{v}}{\mathrm{r}}\\ \mathrm{mgd}\mathrm{²õ¾\pm ²Ôθ}& =& \frac{1}{2}{\mathrm{\pm õÓ\neg }}^2\left(\frac{{\mathrm{mr}}^2}{\mathrm{I}}+1\right)\\ \mathrm{mgd}\mathrm{²õ¾\pm ²Ôθ}& =& ∆{\mathrm{K}}_{\mathrm{rot}}\left(\frac{{\mathrm{mr}}^2}{\mathrm{I}}+1\right)\\ ∆{\mathrm{K}}_{\mathrm{rot}}& =& \frac{\mathrm{mgd}\mathrm{²õ¾\pm ²Ôθ}}{\left({\displaystyle \frac{{\mathrm{mr}}^2}{\mathrm{I}}}+1\right)}\\ & =& \frac{\left(10.0\mathrm{kg}\right)×\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(2.00\mathrm{m}\right)\sin \left(30.0^{°}\right)}{\left[{\displaystyle \frac{\left(10.0\mathrm{kg}\right){\left(0.200\mathrm{m}\right)}^2}{0.600\mathrm{kg}.{\mathrm{m}}^2}}+1\right]}\\ ∆{\mathrm{K}}_{\mathrm{rot}}& =& 58.8\mathrm{J}\end{array}$.

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.