(B) Using the $3^{rd}$ equation of motion: $v^2 = u^2 + 2as$.Since the motion starts from rest,$u = 0$,so $v^2 = 2 a_t s$.Therefore,$a_t = \frac{v^2}{

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(B) Using the $3^{rd}$ equation of motion: $v^2 = u^2 + 2as$.Since the motion starts from rest,$u = 0$,so $v^2 = 2 a_t s$.Therefore,$a_t = \frac{v^2}{2s}$.By the end of the $3^{rd}$ revolution,the total distance covered is $s = 3 \times (2 \pi r) = 6 \pi r$.Substituting $s$ into the acceleration formula: $a_t = \frac{v^2}{2 \times 6 \pi r} = \frac{v^2}{12 \pi r}$.Given the kinetic energy $E = \frac{1}{2} m v^2$,we have $v^2 = \frac{2E}{m}$.Substituting $v^2$ into the expression for $a_t$: $a_t = \frac{2E}{m \times 12 \pi r} = \frac{E}{6 \pi rm}$.

Class 11 Physics 3-2.Motion in Plane Kinematics Circular Motion (Uniform Angular Accelaration)

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$A$ particle of mass $m$ moves along a circle of radius $r$ with constant tangential acceleration. If the kinetic energy $(K.E.)$ of the particle is $E$ by the end of the third revolution after the beginning of the motion,then the magnitude of the tangential acceleration is:

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A
$\frac{E}{2 \pi rm}$
B
$\frac{E}{6 \pi rm}$
C
$\frac{E}{8 \pi rm}$
D
$\frac{E}{4 \pi rm}$

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3-2.Motion in Plane

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Kinematics Circular Motion (Uniform Angular Accelaration)

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$A$ particle of mass $m$ moves along a circle of radius $r$ with constant tangential acceleration. If the kinetic energy $(K.E.)$ of the particle is $E$ by the end of the third revolution after the beginning of the motion,then the magnitude of the tangential acceleration is:

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