Conservation of Energy in Rolling Motion Video Lessons

Concept

# Problem: A solid sphere of mass 2.5kg and radius 0.15m starts from rest at the top of a ramp inclined 25 degrees and rolls to the bottom without slipping. The upper end of the ramp is 0.85m higher than the lower end. a. What is the translational kinetic energy of the sphere when it reaches the bottom of the ramp? b. What is the rotational kinetic energy of the sphere when it reaches the bottom of the ramp?

###### FREE Expert Solution

Potential energy:

$\overline{){\mathbf{P}}{\mathbf{E}}{\mathbf{=}}{\mathbf{m}}{\mathbf{g}}{\mathbf{h}}}$

Kinetic energy:

$\overline{){\mathbf{K}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{{\mathbf{mv}}}^{{\mathbf{2}}}}$

The kinetic energy at the bottom:

$\overline{){{\mathbf{K}}}_{\mathbf{b}\mathbf{o}\mathbf{t}\mathbf{t}\mathbf{o}\mathbf{m}}{\mathbf{=}}{{\mathbf{K}}}_{\mathbf{t}\mathbf{r}\mathbf{a}\mathbf{n}\mathbf{s}\mathbf{l}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}\mathbf{a}\mathbf{l}}{\mathbf{+}}{{\mathbf{K}}}_{\mathbf{r}\mathbf{o}\mathbf{t}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}\mathbf{a}\mathbf{l}}}$

Translational kinetic energy is given by:

Rotational kinetic energy is given by:

$\overline{){{\mathbf{K}}}_{{\mathbf{rotational}}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{I}}{{\mathbf{\omega }}}^{{\mathbf{2}}}}$

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###### Problem Details

A solid sphere of mass 2.5kg and radius 0.15m starts from rest at the top of a ramp inclined 25 degrees and rolls to the bottom without slipping. The upper end of the ramp is 0.85m higher than the lower end.

a. What is the translational kinetic energy of the sphere when it reaches the bottom of the ramp?

b. What is the rotational kinetic energy of the sphere when it reaches the bottom of the ramp?