Question

Circular disc rolls down an inclined plane. The ratio of rotational kinetic energy to total kinetic energy is​

Answer 1

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Answer 2

Given :

Circular disc rolls down an inclined plane .

To Find :

The ratio of rotational kinetic energy to total kinetic energy .

Solution :

We know , kinetic energy is given by :

$K.E=\dfrac{M v^2}{2}$

Rotational energy id given by :

$R.E=\dfrac{I\omega^2}{2}$

Here , $I=\dfrac{MR^2}{2}$ and $v=\omega R$ .

Therefore , $R.E=\dfrac{Mv^2}{4}$

So , total energy is given by :

$T.E=K.E+R.E\\\\T.E=\dfrac{Mv^2}{2}+\dfrac{Mv^2}{4}\\\\T.E=\dfrac{3Mv^2}{4}$

So , ratio of rotational kinetic energy to total kinetic energy is :

$R=\dfrac{R.E}{T.E}\\\\R=\dfrac{\dfrac{Mv^2}{4}}{\dfrac{3Mv^2}{4}}\\\\\\R=\dfrac{1}{3}$

Therefore , the ratio of rotational kinetic energy to total kinetic energy is​ $\dfrac{1}{3}$ .

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