Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 8 - Potential Energy and Conservation of Energy - Problems - Page 203: 17c


30 cm

Work Step by Step

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.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.