Answer:
[tex]v=\sqrt{\dfrac{10g(R-r)}{7}}[/tex]
Explanation:
Given that
Radius of track = R
Radius of ball = r
The ball can be treated as solid sphere, so
The moment of inertia of ball
[tex]I=\dfrac{2}{5}mr^2[/tex]
When the ball reach at the lowest position then it will have both angular and linear speed.
Condition for rolling without slipping v= ωr
Form energy conservation
[tex]mgR=mgr+\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2[/tex]
v= ωr
[tex]I=\dfrac{2}{5}mr^2[/tex]
[tex]mgR=mgr+\dfrac{1}{2}mv^2+\dfrac{1}{2}\times \dfrac{2}{5}mr^2\omega^2[/tex]
[tex]mg(R-r)=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times \dfrac{2}{5}mv^2[/tex]
[tex]2mg(R-r)=mv^2+\dfrac{2}{5}mv^2[/tex]
[tex]2g(R-r)=\dfrac{7}{5}v^2[/tex]
[tex]v=\sqrt{\dfrac{10g(R-r)}{7}}[/tex]
