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A thin walled hollow cylinder is rolling down an incline, without slipping. At any instant, without slipping. At any instant, the ratio "Rotational K.E.: Translational K.E.: Total K.E." is ______. - Physics

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प्रश्न

A thin walled hollow cylinder is rolling down an incline, without slipping. At any instant, without slipping. At any instant, the ratio "Rotational K.E.: Translational K.E.: Total K.E." is ______.

विकल्प

  • 1 : 1 : 2

  • 1 : 2 : 3

  • 1 : 1 : 1

  • 2 : 1 : 3

MCQ
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उत्तर

A thin walled hollow cylinder is rolling down an incline, without slipping. At any instant, without slipping. At any instant, the ratio "Rotational K.E.: Translational K.E.: Total K.E." is 1 : 1 : 2.

Explanation:

We know,

Rotational kinetic energy KR = `1/2 "mv"^2 "K"^2/"R"^2`   ...(1)

Translational K. E. KT = `1/2 "mv"^2`   ...(2)

Total kinetic energy K.ET = `1/2 "mv"^2 (1 + "K"^2/"R"^2)`   ...(3)

`I = mR^2`

`m K^2 = m R^2`

`K^2/R^2 = 1`   ...(4)

∴ K: K: K.E= `1/2 "mv"^2 "K"^2/"R"^2 : 1/2 "mv"^2 : 1/2 "mv"^2 (1 + "K"^2/"R"^2)`

∴ K: K: K.E= `cancel(1/2 "mv"^2) "K"^2/"R"^2 : cancel(1/2 "mv"^2) : cancel(1/2 "mv"^2) (1 + "K"^2/"R"^2)`

∴ K: K: K.E= `"K"^2/"R"^2 : 1 : (1 + "K"^2/"R"^2)`

 K: K: K.E= 1 : 1 : (1 + 1) (From equation 4)

K: K: K.E= 1 : 1 : 2

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Rotational Dynamics
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अध्याय 1: Rotational Dynamics - Exercises [पृष्ठ २४]

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बालभारती Physics [English] Standard 12 Maharashtra State Board
अध्याय 1 Rotational Dynamics
Exercises | Q 1.6 | पृष्ठ २४

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