When a hollow sphere is rolling without slipping on a rough horizontal surface then the percentage of its total K.E which is Translational is
Answers
Solution :
⏭ Given:
✏ A hollow sphere is rolling without slipping on a rough horizotal surface.
⏭ To Find:
✏ The percentage of its total KE which is translation is
⏭ Concept:
✏ Total kinetic energy associated with rolling body (without slipping) is given by
⏭ Calculation:
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Solution:-
Let us consider a sphere of mass is M and radius is R.
★ Using formula:-
Momentum of innertia, I = ⅔ MR²
★ Point to remember:-
Here , the hollow sphere is rolling without slipping.
Therefore, v = Rω.
- Here, v is the speed of centre of mass and ω is the angular speed.
★ Formulas to be noted:-
→ Transitional kinetic energy = ½ Mv²
→ Rotational kinetic energy = ½ Iω²
★ ½ Iω²
½ × (⅔ MR²) × (v/R²) = ⅓ Mv²
→ Total kinetic energy = ½ Mv² + ⅓Mv² = 5/6 Mv²
★ To find % of transitional energy:-
% of transitional energy = ½ Mv²) / (5/6 Mv²) × 100 = 60%
Hence, the percentage of total kinetic energy which is transitional is 60%.