Question

A Uniform Solid Cylinder Rolls Without Slipping Down An Incline. At The Bottom Of The Incline, The Speed V Of Cylinder Is Measured And The Translation And Rotational Kinetic Energies Are Calculated. A Hole Is Drilled Through The Cylinder Along Its Axis And The Experiment Is Repeated; At The Bottom Of The Incline The Cylinder Now Has Translation And Rotational Kinetic Energies How Does The Ratio Of Rotational To Translational Kinetic Energy Of The Cylinder Compare To Its Original Value?

Answer 1

Answer:

A value chain consisting of two types of additional storage in the matter of time

Answer 2

Ratio Of Rotational To Translational Kinetic Energy Of The Cylinder Compare To Its Original Value is $2:3$.

Explanation:

$k_{1} = k$ × $E$ of translation.

$E =$ total  $K$ × $E$ energy of solid cylinder.

Now, $\frac{K_{1} }{E} = ?$.

$E = K_{1} + K_{-2}$

$K_{2t}$   ( $K$ × $E$ of $R_{0+n}$ )

So, $K_{1} = \frac{1}{2} mv^{2}$ and $K_{2} = \frac{1}{2}IW^{2}$.

$E = \frac{1}{2} mv^{2} + \frac{1}{2}$ × $\frac{MR^{2} }{2}$ × $(\frac{V^{2} }{R^{2} } )$

$E = \frac{1}{2} mv^{2} + \frac{mv^{2} }{4} = \frac{3mv^{2} }{4}$

$\frac{K_{1} }{E} = \frac{\frac{1}{2}mv^{2} }{\frac{3}{4} mv^{2} } = \frac{1}{2}$ × $\frac{4}{3} = \frac{2}{3}$

∴ $\frac{K_{1} }{E} = \frac{2}{3}$

So, the Ratio Of Rotational To Translational Kinetic Energy Of The Cylinder Compare To Its Original Value is $2:3$.

#SPJ2