Question 300: Work, Energy and Power & Circular Motion

Question

A particle of mass $$m$$ is performing UCM along a circle of radius $$r$$. The relation between centripetal acceleration $$a$$ and kinetic energy $$E$$ is given by

Accepted Answer

To find the relation between centripetal acceleration $$a$$ and kinetic energy $$E$$ for a particle of mass $$m$$ performing uniform circular motion of radius $$r$$, we start by expressing the centripetal acceleration and the kinetic energy in terms of the velocity $$v$$ of the particle.
The centripetal acceleration $$a$$ is given by:
$$ a = \frac{v^2}{r} $$
The kinetic energy $$E$$ of the particle is given by:
$$ E = \frac{1}{2} m v^2 $$
From the expression for kinetic energy, we can solve for $$v^2$$:
$$ v^2 = \frac{2E}{m} $$
Substitute $$v^2$$ from the kinetic energy expression into the centripetal acceleration expression:
$$ a = \frac{v^2}{r} = \frac{\frac{2E}{m}}{r} = \frac{2E}{m r} $$
Therefore, the correct relation between centripetal acceleration $$a$$ and kinetic energy $$E$$ is:
Option A
$$ a = \frac{2 E}{m r} $$

Answer

$$a=\frac{2 E}{m r}$$

Answer

$$a=2 E m$$

Answer

$$a=\frac{E}{m r}$$

Answer

$$a=\left(\frac{2 E}{m r}\right)^2$$

Work, Energy and Power & Circular Motion

A particle of mass $m is performing UCM along a circle of radius r. The relation between centripetal acceleration a and kinetic energy E$ is given by

Choose the correct answer: