## Fundamentals of Physics Extended (10th Edition)

If the normal force is 0, it means that the entire centripetal force $F_{c} = \frac {mv^{2}}{R}$ has to be provided by the gravitational force $F_{g} = mg$, which is the only other force acting on the block. In Problem 17a, we used $F_{c} = \frac {m(v_{0}^{2} – 2g\Delta h)}{R}$ to find the centripetal force. We thus have that $F_{c} = F_{g}$ $\frac {m(v_{0}^{2} – 2g\Delta h)}{R} = mg$ Solving for $\Delta h$: $\frac {m(v_{0}^{2} – 2g\Delta h)}{R} = mg$ $v_{0}^{2} – 2g\Delta h = gR$ $– 2g\Delta h = – v_{0}^{2} + gR$ $\Delta h = \frac{v_{0}^{2} – gR}{2g}$ Plugging in our know values: $\Delta h = \frac{0^{2} – (9.8 m/s²)(0.12 m)}{2(9.8 m/s²)}$ $\Delta h$ = –0.06 m This means that the vertical displacement from where the block is released to its position at the top of the loop has to be –0.06 m. The top of the loop is 2R = 2(0.12 m) = 0.24 m above the bottom of the loop, so for it to have this $\Delta h$ it must have been released from a height $h$ = 0.24 m + 0.06 m = 0.30 m = 30 cm.