Q.What is expected to happen if one travels fast over a speed breaker? Why?

Vertical circular motion controlled purely by gravity (after an initial push) is genuinely different from uniform circular motion, because the speed keeps changing throughout the motion -- fastest at the bottom, slowest at the top -- so the resultant force is centripetal only at the very top and bottom of the circle; everywhere else, part of the resultant is tangential, changing the speed rather than the direction. For a bob on a string (radius r = string length), the minimum speeds required at the top and bottom for the string to stay taut are $v_{A,min}=\sqrt{rg}$ (top) and $v_{B,min}=\sqrt{5rg}$ (bottom), related by energy conservation ($v_B^2-v_A^2=4gr$), while at the two positions where the string is exactly horizontal, $v_{C,min}=v_{D,min}=\sqrt{3rg}$.

A particularly elegant result is that the DIFFERENCE between the extreme tensions, $T_B-T_A=6mg$, depends only on the object's weight -- completely independent of the radius or the actual speed involved, since both cancel out in the derivation. Because a string can only pull (never push), it can never allow zero speed at the top (that would require negative tension); this is only possible for the alternative case of a rigid ROD, which can push as well as pull. None of the ordinary kinematical equations of motion apply during this non-uniform motion, since neither the linear nor angular acceleration stays constant -- only energy conservation and the force equations at specific points can be used.