(A) The normal acceleration is given by $a_n = \frac{v^2}{R} = k t^\alpha$, where $k$ is a constant.From this, the square of the velocity is $v^2 = R

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(A) The normal acceleration is given by $a_n = \frac{v^2}{R} = k t^\alpha$, where $k$ is a constant.From this, the square of the velocity is $v^2 = R k t^\alpha$.The kinetic energy of the object is $K = \frac{1}{2} m v^2 = \frac{1}{2} m R k t^\alpha$.The power $P$ developed by all forces is equal to the rate of change of kinetic energy, $P = \frac{dK}{dt}$.$P = \frac{d}{dt} \left( \frac{1}{2} m R k t^\alpha \right) = \frac{1}{2} m R k \alpha t^{\alpha-1}$.Since $m$, $R$, $k$, and $\alpha$ are constants, we have $P \propto t^{\alpha-1}$.

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

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An object moves along a circle with normal acceleration proportional to $t^\alpha$, where $t$ is the time and $\alpha$ is a positive constant. The power developed by all the forces acting on the object will have time dependence proportional to

TS EAMCET 2018 All TS EAMCET

A
$t^{\alpha-1}$
B
$t^{\alpha / 2}$
C
$t^{\frac{1+\alpha}{2}}$
D
$t^{2 \alpha}$

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3-2.Motion in Plane

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Kinematics Circular Motion (Uniform Angular Accelaration)

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An object moves along a circle with normal acceleration proportional to $t^\alpha$, where $t$ is the time and $\alpha$ is a positive constant. The power developed by all the forces acting on the object will have time dependence proportional to

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