2 A stone is released from rest at a height h at the left si

2) A stone is released from rest at a height h at the left side of a loop-the-loop, as shown in (2pts) the figure. There is no appreciable friction from the track or from air resistance. If the radius of the loop is R, what is the maximum radius R for which the stone will not fall off the track at the top of the loop? to 2.5 h 0.4 h 0.25 h 0.5 h

Solution

the minimum speed of stone needed at topmost point of loop of radius R is given by

mv²/R = mg

v=sqrt(gR)

Total energy of stone at topmost point of loop of radiusR = m*g*2R + 0.5*m*[sqrt(gR)]² = 2.5mgR

h should be such that the potential energy of stone =2.5mgR

mgh=2.5mgR

h = 2.5R

R=h/2.5 = 0.4h

Fouth option is correct