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Question
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 ______.
Options
1 : 1 : 2
1 : 2 : 3
1 : 1 : 1
2 : 1 : 3
Solution
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)
∴ KR : KT : K.ET = `1/2 "mv"^2 "K"^2/"R"^2 : 1/2 "mv"^2 : 1/2 "mv"^2 (1 + "K"^2/"R"^2)`
∴ KR : KT : K.ET = `cancel(1/2 "mv"^2) "K"^2/"R"^2 : cancel(1/2 "mv"^2) : cancel(1/2 "mv"^2) (1 + "K"^2/"R"^2)`
∴ KR : KT : K.ET = `"K"^2/"R"^2 : 1 : (1 + "K"^2/"R"^2)`
KR : KT : K.ET = 1 : 1 : (1 + 1) (From equation 4)
KR : KT : K.ET = 1 : 1 : 2
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