Rotational Motion Question 79
Question: A small object of uniform density rolls up a curved surface with an initial velocity v’. It reaches up to a maximum height of $ \frac{3{{\upsilon }^{2}}}{4g} $ with respect to the initial position. The object is
Options:
A) ring
B) solid sphere
C) hollow sphere
D) disc
Show Answer
Answer:
Correct Answer: D
Solution:
-
As $ v=\sqrt{\frac{2gh}{1+\frac{k^{2}}{r^{2}}}} $
Given $ h=\frac{3v^{2}}{4g} $ $ v^{2}=\frac{2gh}{1+\frac{k^{2}}{r^{2}}}=\frac{2g3v^{2}}{4g( 1+\frac{k^{2}}{r^{2}} )}=\frac{6gv^{2}}{4g( 1+\frac{u^{2}}{v^{2}} )} $
$ 1=\frac{3}{2( 1+\frac{k^{2}}{v^{2}} )} $ or $ 1+\frac{k^{2}}{r^{2}}=\frac{3}{2} $ or $ \frac{k^{2}}{r^{2}}=\frac{3}{2}-1=\frac{1}{2} $ $ k^{2}=\frac{1}{2}r^{2} $
(Equation of disc) Hence, the object is disc.
