(A) The kinetic energy $(E)$ of a particle of mass $m$ moving with speed $v$ is given by $E = \frac{1}{2} m v^2$.For uniform circular motion,the centr

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(A) The kinetic energy $(E)$ of a particle of mass $m$ moving with speed $v$ is given by $E = \frac{1}{2} m v^2$.For uniform circular motion,the centripetal acceleration $(a)$ is given by $a = \frac{v^2}{r}$,which implies $v^2 = a r$.Substituting $v^2 = a r$ into the kinetic energy equation:$E = \frac{1}{2} m (a r)$Rearranging the equation to solve for acceleration $(a)$:$2 E = m a r$$a = \frac{2 E}{m r}$

Class 11 Physics 2 Motion in a Plane Angular Variables and Basic of Uniform Circular Motion

Easy

If the kinetic energy $(KE)$ of a particle of mass $m$ performing uniform circular motion $(UCM)$ in a circle of radius $r$ is $E$,find the acceleration of the particle.

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A
$\frac{2 E}{m r}$
B
$\left(\frac{2 E}{m r}\right)^{2}$
C
$2 E m r$
D
$\frac{4 E}{m r}$

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2 Motion in a Plane

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Angular Variables and Basic of Uniform Circular Motion

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If the kinetic energy $(KE)$ of a particle of mass $m$ performing uniform circular motion $(UCM)$ in a circle of radius $r$ is $E$,find the acceleration of the particle.

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$A$ particle $P$ is moving in a circle of radius $a$ with a uniform speed $v$. $C$ is the centre of the circle and $AB$ is a diameter. When passing through $B$,the angular velocity of $P$ about $A$ and $C$ are in the ratio:

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