Conservation of Energy in Rolling Motion Video Lessons

Concept

# Problem: An 8.0 cm diameter, 400 g sphere is released from rest at the tip of a 2.1 m long, 25 degree incline. It rolls, without slipping, to the bottom.What is the sphere's angular velocity at the bottom of the incline?

###### FREE Expert Solution

Gravitational potential energy:

$\overline{){\mathbf{U}}{\mathbf{=}}{\mathbf{m}}{\mathbf{g}}{\mathbf{h}}}$

Moment of inertia of a sphere:

$\overline{){\mathbf{I}}{\mathbf{=}}\frac{\mathbf{2}}{\mathbf{5}}{\mathbf{m}}{{\mathbf{r}}}^{{\mathbf{2}}}}$

Rotational kinetic energy:

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

Translational kinetic energy:

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

We'll use trigonometry to determine the height of the incline.

sinθ = h/L

h = Lsinθ

The gravitational potential energy of the sphere is converted into kinetic energy of the sphere.

U = Krot + Ktranslational

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###### Problem Details

An 8.0 cm diameter, 400 g sphere is released from rest at the tip of a 2.1 m long, 25 degree incline. It rolls, without slipping, to the bottom.

What is the sphere's angular velocity at the bottom of the incline?

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