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 thin wall sphere with a moment of inertia (2/3)mR²

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

mgh = ½mv² + ½Iω²

mgh = ½mv² + ½((2/3)mR²)(v/R)²

2gh = v² + 2/3v²

2gh = 5/3v²

v = √((6/5)gh)

v = √((6/5)(9.81)(10.0sin30.0))

A) v = 7.67 m/s  

B) ω = 7.67/0.345 = 22.2 radians/s

C)  ½mv² / ½Iω² = mv²/Iω² = mv²/((2/3)mR²)(v/R)² = 3/2

newton's first law of motion

the moving of the box is a result of centripetal force. think of that  as a car or a truck in this case creates a turn, the force of friction performing upon the turned wheels of the car runs centripetal force obligatory for circular motion for the things inside it.