Work Energy And Power Question 255
Question: A bead of mass m is sliding down the fixed inclined rod without friction. It is connected to a point P on the horizontal surface with a light spring of spring constant k. The bead is initially released from rest and the spring is initially unstressed and vertical. The bead just stops at the bottom of the inclined rod. Find the angle which the inclined rod makes with horizontal.
Options:
A) $ {{\cot }^{-1}}( 1+\sqrt{\frac{2mg}{kh}} ) $
B) $ ta{n^{-1}}( 1+\sqrt{\frac{2mg}{kh}} ) $
C) $ {{\cot }^{-1}}( 1+\sqrt{\frac{mg}{kh}} ) $
D) $ ta{n^{-1}}( 1+\sqrt{\frac{mg}{kh}} ) $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Loss in gravitational $ P.E.= $ gain in spring RE.
$ mgh=\frac{1}{2}k{{( hcot\alpha -h )}^{2}}or( \cot \alpha -1 )=\sqrt{\frac{2mg}{kh}} $
Or, $ \cot \alpha =1+\sqrt{\frac{2mg}{kh}} $
