Conservation of Energy in Rolling Motion Video Lessons

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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:

U=mgh

Moment of inertia of a sphere:

I=25mr2

Rotational kinetic energy:

Krot=12Iω2

Translational kinetic energy:

K=12mv2

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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