Starting from rest, a solid sphere rolls without slipping down an incline plane. At the bottom of the incline, what does the angular velocity of the sphere depend upon?

2/R*sqrt (g*s*sin (θ)) = w

Explanation:

Assume:

- The cylinder with mass m

- The radius of cylinder R

- Distance traveled down the slope is s

- The angular speed at bottom of slope w

- The slope of the plane θ

- Frictionless surface.

Solution:

- Using energy principle at top and bottom of the slope. The exchange of gravitational potential energy at height h, and kinetic energy at the bottom of slope.

ΔPE = ΔKE

- The change in gravitational potential energy is given as m*g*h.

- The kinetic energy of the cylinder at the bottom is given as rotational motion: 0.5*I*w^2

- Where I is the moment of inertia of the cylinder I = 0.5*m*R^2

We have:

m*g*s*sin (θ) = 0.25*m*R^2*w^2

2/R*sqrt (g*s*sin (θ)) = w

- The angular velocity depends on plane geometry θ, distance travelled down slope s, Radius of the cylinder R, and gravitational acceleration g