A sphere of radius r =34.5 cm and mass m = 1.80 kg starts from rest and rolls without slipping down a 30.0∘ incline that is 10.0 m long. A.) calculate its translational speed when it reaches the bottom.
B.) Calculate its rotational speed when it reaches the bottom.
C.) what is the ratio of translational to rotational kinetic energy at the bottom?

Explanation:

It depends on the construction of the sphere.

Let's ASSUME it is uniform construction with a moment of inertia (2/5)mR²

The Change in potential energy will equal the change in kinetic energy

mgh = ½mv² + ½Iω²

mgh = ½mv² + ½((2/5)mR²)(v/R)²

2gh = v² + 2/5v²

2gh = 7/5v²

v = √((10/7)gh)

v = √((10/7)(9.81)(10.0sin30.0))

A) v = 8.37 m/s  

B) ω = 8.37/0.345 = 24.3 radians/s

C)  ½mv² / ½Iω² = mv²/Iω² = mv²/((2/5)mR²)(v/R)² = 5/2

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.