Rotational Motion Question 138
Question: A small solid sphere of radius $r$ rolls down an incline without slipping which ends into a vertical loop of radius $R$ . Find the height above the base so that it just loops the loop
Options:
A) $\frac{5}{2}R$
B) $\frac{5}{2}(R-r)$
C) $\frac{25}{10}(R-r)$
D) $\frac{27}{10}R-\frac{17r}{10}$
Show Answer
Answer:
Correct Answer: D
Solution:
[d]The minimum velocity at P, top of the loop, should be
$v=\sqrt{g(R-r)}$ , if the sphere keeps on rolling at top
$v=\omega R$
$mgh=\frac{1}{2}mv^{2}+\frac{1}{2}I{{\omega }^{2}}+mg(2R-r)$
$=\frac{1}{2}mg(R-r)+\frac{1}{2}( \frac{2}{5} )mR^{2}2{{\omega }^{2}}+mg(2R-r)$
$=\frac{7}{10}mg(R-r)+mg(2R-r)$
$\omega R=v=\sqrt{g(R-r)} $
$=\frac{mg}{10}(27R-17r)orh=\frac{1}{10}(27R-17r)$
