Answer:
[tex]h>\dfrac{5}{2}R[/tex]
Explanation:
Given that
Height = h
Radius = R
From energy conservation
[tex]KE_A+U_A=KE_B+U_B[/tex]
At point B
The minimum speed to complete the the circle
[tex]V_B=\sqrt{gR}\ m/s[/tex]
So the kinetic energy at point B
[tex]KE_B=\dfrac{1}{2}mV^2[/tex]
[tex]KE_B=\dfrac{1}{2}mgR[/tex]
[tex]KE_A+U_A=KE_B+U_B[/tex]
[tex]0+mgh=\dfrac{1}{2}mgR+2mgR[/tex]
Without falling off at the top (point B)
[tex]0+mgh>\dfrac{1}{2}mgR+2mgR[/tex]
[tex]mg(h-2R)>\dfrac{1}{2}mgR[/tex]
[tex]g(h-2R)>\dfrac{1}{2}gR[/tex]
[tex]h>\dfrac{5}{2}R[/tex]