Answer:
[tex]v=\sqrt{2gR}\ m/s[/tex]
Explanation:
Given that
Mass = m
Height h=R
acceleration due to gravity = g m/s²
Initially the speed of the mass ,u=0 m/s
The final speed of the mass at bottom = v m/s
Now from work power energy theorem
Work done by all forces=Change in the kinetic energy
Given that surface is friction less that is why work done by the friction force is zero.
[tex]mgh=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2[/tex]
[tex]mgR=\dfrac{1}{2}mv^2[/tex]
[tex]v=\sqrt{2gR}\ m/s[/tex]
Therefore the speed at the bottom of the circular loop is [tex]\sqrt{2gR}[/tex]
