Kinematics Question 484
Question: The kinetic energy $ k $ of a particle moving along a circle of radius $ R $ depends on the distance covered $ s $ as $ k=as^{2} $ where $ a $ it’s a constant. The force acting on the particle is [MNR 1992; JIPMER 2001, 02; AMU (Engg.) 1999]
Options:
A) $ 2a\frac{s^{2}}{R} $
B) $ 2as{{( 1+\frac{s^{2}}{R^{2}} )}^{1/2}} $
C) $ 2as $
D) $ 2a\frac{R^{2}}{s} $
Show Answer
Answer:
Correct Answer: B
Solution:
According to given problem $ \frac{1}{2}mv^{2}=as^{2} $
$ \Rightarrow v=s\sqrt{\frac{2a}{m}} $
So $ a _{R}=\frac{v^{2}}{R}=\frac{2as^{2}}{mR} $ -(i)
Further more as $ a _{t}=\frac{dv}{dt}=\frac{dv}{ds}\cdot \frac{ds}{dt}=v\frac{dv}{ds} $ -(ii)
(By chain rule) Which in light of equation (i) i.e. $ v=s\sqrt{\frac{2a}{m}} $
yields $ a _{t}=[ s\sqrt{\frac{2a}{m}} ][ \sqrt{\frac{2a}{m}} ]=\frac{2as}{m} $ -(iii)
So that $ a=\sqrt{a _{R}^{2}+a _{t}^{2}}=\sqrt{{{[ \frac{2as^{2}}{mR} ]}^{2}}+{{[ \frac{2as}{m} ]}^{2}}} $
Hence $ a=\frac{2as}{m}\sqrt{1+{{[s/R]}^{2}}} $
$ F=ma=2as\sqrt{1+{{[s/R]}^{2}}} $
