Answer:
v = 11.4 m/s
Explanation:
- The only force that keeps the object car+rider on the loop, at any point, is the centripetal force.
- This force always points inward to the center of the loop.
- This force is not a different type of force, is just the net force in this direction at any point.
- At the top of the loop, it's pointing just downward, and it's the sum of the weight and the normal force (both pointing downward).
- In the limit case, when the rider is on the verge of falling out, the normal force is just zero, so the centripetal force must be equal to the weight, as follows:
[tex]F_{g} = F_{cent}[/tex]
- Replacing both forces by their expressions, we get:
[tex]m*g = \frac{m*v^{2}}{r} (1)[/tex]
- Simplifying, rearranging and solving for v, we get:
[tex]v_{min} = \sqrt{r*g} = \sqrt{13.2m * 9.8 m/s2} = 11.4 m/s[/tex]
- If v> vmin, the centripetal force will be higher, so there must be a normal force added to the weight, to keep the equation (1) balanced.
