Answer:[tex]v=\sqrt{\frac{10}{7}\cdot gH}[/tex]
Explanation:
Given
Mass of ball is M
Height of incline is H
Here conservation of energy will provide the velocity at bottom
Energy at top of incline plane [tex]=MgH[/tex]
Energy at bottom=Kinetic energy+Rotational energy
Assuming Pure rolling we can write
[tex]v=\omega R[/tex]
where [tex]v[/tex]=velocity of ball
[tex]\omega [/tex]=angular velocity of ball
R=radius of ball
[tex]E_b=\frac{1}{2}Mv^2+\frac{1}{2}I\omega ^2[/tex]
where I=moment of inertia of ball
[tex]I=\frac{2}{5}MR^2[/tex]
[tex]E_b=\frac{1}{2}Mv^2+\frac{1}{2}\times \frac{2}{5}MR^2\times (\frac{V}{R})^2[/tex]
[tex]E_b=\frac{7}{10}Mv^2[/tex]
Now ,
[tex]MgH=\frac{7}{10}Mv^2[/tex]
[tex]v=\sqrt{\frac{10}{7}\cdot gH}[/tex]
