Question

The kinetic energy K of a particle of mass m moving along a circle of radius R depends on distance covered s as K= as2. Then the acceleration of particle is given by

• A. $ \frac{2as}{m}{{\left( 1+\frac{{{s}^{2}}}{{{R}^{2}}} \right)}^{1/2}} $
• B. $ \frac{2as}{m}{{\left( 1-\frac{{{s}^{2}}}{{{R}^{2}}} \right)}^{1/2}} $
• C. $ \frac{2a{{s}^{2}}}{mR} $
• D. $ \frac{2as}{m} $

Answer 1

The Correct Option is A

According to given problem $ \frac{1}{2}m{{v}^{2}}=a{{s}^{2}} $ $ v=s\sqrt{\frac{2a}{m}} $ So $ {{a}_{R}}=\frac{{{v}^{2}}}{R}=\frac{2a{{s}^{2}}}{mR} $ Furthermore as $ {{a}_{t}}=\frac{dv}{dt}=\frac{dv}{ds}\times \frac{ds}{dt}=v\frac{dv}{ds} $ $ {{a}_{t}}=\left[ s\sqrt{\frac{2a}{m}} \right]\left[ \sqrt{\frac{2a}{m}} \right] $ $ \left[ \because v=s\sqrt{\frac{2a}{m}}\text{ and }\frac{dv}{ds}=\sqrt{\frac{2a}{m}} \right] $ $ {{a}_{t}}=\frac{2as}{m} $ Acceleration $ a=\sqrt{a_{R}^{2}+a_{t}^{2}} $ $ =\sqrt{\left[ \frac{2a{{s}^{2}}}{mR} \right]+{{\left[ \frac{2as}{m} \right]}^{2}}} $ $ =\frac{2as}{m}\sqrt{\left( 1+\frac{{{s}^{2}}}{{{R}^{2}}} \right)} $