Q134 / 180 · COMEDK 2022 · Physics
A particle of mass $m$ is moving in a horizontal circle of radius $r$ under a centripetal force given by $\left(\dfrac{-K}{r^2}\right)$, where $K$ is a constant. Then
- Correct answerAthe total energy of the particle is $\left(\dfrac{-K}{2 r}\right)$
- Bthe kinetic energy of the particle is $\left(\dfrac{K}{r}\right)$
- Cthe potential energy of the particle is $\left(\dfrac{K}{2 r}\right)$
- Dthe kinetic energy of the particle is $\left(\dfrac{-K}{r}\right)$
Solution
The centripetal force acting on the particle is given by $F = \dfrac{-K}{r^2}$. The magnitude of the centripetal force is $F_c = \dfrac{mv^2}{r} = \dfrac{K}{r^2}$.
From this, the kinetic energy $K.E.$ is calculated as $K.E. = \dfrac{1}{2}mv^2 = \dfrac{1}{2} \left( \dfrac{K}{r} \right) = \dfrac{K}{2r}$.
The potential energy $U$ is defined by $F = -\dfrac{dU}{dr}$. Integrating the force $F = -\dfrac{K}{r^2}$ with respect to $r$, we get $U = \int \dfrac{K}{r^2} dr = -\dfrac{K}{r}$.
The total energy $E$ is the sum of kinetic energy and potential energy: $E = K.E. + U = \dfrac{K}{2r} - \dfrac{K}{r} = -\dfrac{K}{2r}$.
Comparing this with the given options, the total energy of the particle is $-\dfrac{K}{2r}$.
Answer: the total energy of the particle is $\left(\dfrac{-K}{2 r}\right)$
From this, the kinetic energy $K.E.$ is calculated as $K.E. = \dfrac{1}{2}mv^2 = \dfrac{1}{2} \left( \dfrac{K}{r} \right) = \dfrac{K}{2r}$.
The potential energy $U$ is defined by $F = -\dfrac{dU}{dr}$. Integrating the force $F = -\dfrac{K}{r^2}$ with respect to $r$, we get $U = \int \dfrac{K}{r^2} dr = -\dfrac{K}{r}$.
The total energy $E$ is the sum of kinetic energy and potential energy: $E = K.E. + U = \dfrac{K}{2r} - \dfrac{K}{r} = -\dfrac{K}{2r}$.
Comparing this with the given options, the total energy of the particle is $-\dfrac{K}{2r}$.
Answer: the total energy of the particle is $\left(\dfrac{-K}{2 r}\right)$
Hence, the correct answer is option: 'A'.