Answer:
The speed of the car at the apex of the loop must be grater than 2.45 m/s
Explanation:
In order for the car to not fall off the track at the apex of the loop, the norm force of the track at the apex must be greater than zero.
Assuming frictionless life on the track which is also to have a perfectly circular shape near the top (radius being 0.25m), the norm force of the track and gravity both point down and result in the centripetal force:
[tex]F_c=F_N+F_g[/tex]
The formula for centripetal force on a circular trajectory is
[tex]F_c = m\frac{v^2}{r}[/tex]
and so the condition for the car to stay on the track can be written as
[tex]m\frac{v^2}{r} = F_N + mg\implies F_N = m\frac{v^2}{r}-mg>0\\\implies |v| >\sqrt{gr}=\sqrt{9.8\frac{m}{s^2}0.25m}=2.45\frac{m}{s}[/tex]
The speed of the car at the apex of the loop must be grater than 2.45 m/s