Respuesta :
Answer:1
Explanation:
Thin-walled hollow tube is rolling without sliding on the floor
Moment of inertia of the hollow cylinder [tex]I=mr^2[/tex]
where m=mass of cylinder
r=radius of cylinder
Rotational Energy of cylinder[tex]=\frac{1}{2}I\omega^2[/tex]
In rolling [tex]\omega =\frac{v}{r}[/tex]
[tex]R.K.E=\frac{1}{2}\times mr^2\times (\frac{v}{r})^2[/tex]
[tex]R.K.E.=\frac{1}{2}mv^2[/tex]
Translational kinetic Energy of Cylinder is given by
[tex]T.K.E.=\frac{1}{2}mv^2[/tex]
Ratio of translational kinetic energy to rotational Kinetic Energy
[tex]=\frac{\frac{1}{2}mv^2}{\frac{1}{2}mv^2}[/tex]
[tex]=1[/tex]
