Physics, asked by malimukesh8875, 11 months ago

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

Answered by Anonymous
23

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

 \bigstar \sf \:  \red{KE_{net} = KE_{Translation} + KE_{Rotational}} \\  \\  \bigstar \:  \underline{ \boxed{ \bold{ \sf{ \purple{ \large{ \large{KE_{net} =  \dfrac{1}{2} M {v}^{2} { \huge{(}}1 +  \dfrac{ {K}^{2} }{ {R}^{2} }{ \huge{) }}}}}}}}}

Calculation:

 \implies \sf \: for \: hollow \: sphere \\  \\  \dag \sf \: I = M {K}^{2}  =  \dfrac{2}{3} M {R}^{2}  \\  \\  \dag  \:  \boxed{\sf { \blue{ \dfrac{ {K}^{2} }{ {R}^{2} }  =  \dfrac{2}{3} }}} \:  \circ

_________________________________

 \mapsto \sf \: \%(KE)_{Translation} =  \dfrac{KE_{Translation}}{KE_{net}}  \times 100 \\  \\  \mapsto \sf \: \%(KE)_{Translation} =  \dfrac{  \cancel{\dfrac{1}{2} M {v}^{2} }}{  \cancel{\dfrac{1}{2} M {v}^{2}} (1 +  \dfrac{ {K}^{2} }{ {R}^{2} } )}  \times 100 \\  \\  \mapsto \sf \: \%(KE)_{Translation} =  \dfrac{1}{1 +  \dfrac{2}{3} }  \times 100 \\  \\  \mapsto \sf \: \%(KE)_{Translation} =  \frac{3}{5}  \times 100 \\  \\  \mapsto \:  \underline{ \boxed{ \bold{ \sf{ \orange{ \large{\%(KE)_{Translation} = 60\% \: of \: KE_{net}}}}}}} \:  \green{ \bigstar}

Answered by Anonymous
14

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 + Mv² = 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%.

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