Rolling Kinetic Energy

Classical Mechanics Level 2

A disc and a ring of the same mass are rolling without slipping. If their kinetic energies are equal, then what is the ratio of the velocities of their centers?

3 : 4 \sqrt 3 :\sqrt 4 4 : 3 \sqrt 4 :\sqrt 3 2 : 3 \sqrt 2 :\sqrt 3 3 : 2 \sqrt 3 :\sqrt 2

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Rohit Gupta by Rohit Gupta , 34, India

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1 solution

Akhil Bansal
Feb 10, 2016

Let v 1 v_1 and v 2 v_2 be the speed of disc and ring respectively and R 1 R_1 and R 2 R_2 be the radius of disc and ring respectively.

⇒ K E disc = K E ring \Rightarrow \large KE_{\text{disc}} = KE_{\text{ring}} 1 2 m v 1 2 + 1 2 I 1 ω 2 = 1 2 m v 2 2 + 1 2 I 2 ω 2 \large \dfrac{1}{2}mv_1^2 + \dfrac{1}{2}I_1\omega^2 = \dfrac{1}{2}mv_2^2 + \dfrac{1}{2}I_2\omega^2 1 2 m v 1 2 + 1 2 ⋅ m R 1 2 2 ⋅ v 1 2 R 1 2 = 1 2 m v 2 2 + 1 2 ⋅ m R 2 2 ⋅ v 2 2 R 2 2 \large \dfrac{1}{2}mv_1^2 + \dfrac{1}{2} \cdot \dfrac{mR_1^2}{2} \cdot \dfrac{v_1^2}{R_1^2} = \dfrac{1}{2}mv_2^2 + \dfrac{1}{2} \cdot mR_2^2 \cdot \dfrac{v_2^2}{R_2^2} v 1 v 2 = 4 3 \large \dfrac{v_1}{v_2} = \sqrt{\dfrac{4}{3}}

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What is R here

Bishwa Roy - 5 years, 4 months ago

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R is Radius of the disc and ring . The mRR is the equation of inertia, btw

David Groch - 5 years, 4 months ago

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