A circular disc rolls down on an inclined plane without slipping. what fraction of totoal

A circular disc rolls down on an inclined plane without slipping. what fraction of totoal energy is translational?

2 Answers

Consider the general case....

K.E._rotational = 12 I w²

K.E _{translational} = 12 M v²

since pure rolling....so v=wR

so K.E _{translational} = 12 M w² R²

frac of translational energy = K.E _{translational}K.E _{translational}+K.E _rotational

plugging in the values.....

frac of translational energy = 12 M w² R²12 M w² R² + 12 I w²

so we get....

frac K.E. trans... = M R²MR² + I

or .. . =11+I/(MR²)

I = MK²

K = radius of gyration....

so we get.. = 11+MK²/MR²

so K.E._{frac. trnsln} = R²R² + K²

so K.E._{frac. rotnl} = K²R² + K²
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thus in case of a circular disc K² = R²/2

so
so K.E._{frac. trnsln} = R²R² + R²/2 = 23

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for disc K_RK_T = 12

K_R = 12IÏ‰² = \frac{1}{2}\left(\frac{1}{2}mR^{2} \right)\left(\frac{v}{R} \right)^{2} = \frac{1}{4}mv^{2}
K_T = 12mv²

hence K_RK_T = 12
K_RK_T + 1 = 32
K_TK_T+K_R = 23

hence fraction of totoal energy is translational is 23

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