Rotational Motion Question 89
Question: A body initially at rest and sliding along a frictionless track from a height h just completes a vertical circle of diameter $ \text{AB=D} $ . The height h is equal to
Options:
A) $ \frac{7}{5}D $
B) $ D $
C) $ \frac{3}{2}D $
D) $ \frac{5}{4}D $
Show Answer
Answer:
Correct Answer: D
Solution:
-
[d]
As track is frictionless, so total mechanical energy will remain constant
$ T\text{.M}\text{.}{E_l}\text{=T}\text{.M}\text{.}{E_F} $
$ \text{0+mgh=}\frac{1}{2}mv _L^{2}+0 $
$ h=\frac{v_L^{2}}{2g} $
For completing the vertical circle, $ v _{L}\ge \sqrt{5gR} $
$ h=\frac{5gR}{2g}=\frac{5}{2}R=\frac{5}{4}D $
