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

Answer: (A)

$\frac{2as}{m}{{\left( 1+\frac{{{s}^{2}}}{{{R}^{2}}} \right)}^{1/2}}$

Answer 2

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 $a=\sqrt{a_{R}^{2}+a_{t}^{2}}$ $=\sqrt{\left[ \frac{2a{{s}^{2}}}{mR} \right]+{{\left[ \frac{2as}{m} \right]}^{2}}}$ $=\frac{2as}{m}\sqrt{\left( 1+\frac{{{s}^{2}}}{{{R}^{2}}} \right)}$