Answer:
in order to be safe the centripetal acceleration should be higher than 39.2 m/s²
Step-by-step explanation:
Assuming no friction while travel around the loop then we can apply the conservation of energy
where
K = kinetic energy = 1/2*m*v²
and m= mass , v= velocity at the bottom
K = kinetic energy = m*g*h , where g= gravity and h= height from the bottom of the loop
then by conservation of energy
Energy in the bottom of the loop = K + V(=0 because of h=0) = K = (1/2)*m*v²
Energy in the top of the loop = K(=0 in the limiting case of v=0 , that is falling from the loop) + V= V = m*g* 2R
where R= radius of the loop
then
Energy in the bottom of the loop= Energy in the top of the loop
(1/2)*m*v²= m*g* 2R
(1/2)*v²= 2*g* R
v²/R = 4*g
the centripetal acceleration ac is defined as
ac= v²/R= 4*g
then since g=9.8 m/s²
ac= 4*9.8 m/s² = 39.2 m/s²
therefore in order to be safe the centripetal acceleration should be higher than 39.2 m/s²