(B) The centripetal acceleration is given by $a_c = k^2rt^2$. Since $a_c = \frac{v^2}{r}$,we have $\frac{v^2}{r} = k^2rt^2$. Solving for $v$,we get $v

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(B) The centripetal acceleration is given by $a_c = k^2rt^2$. Since $a_c = \frac{v^2}{r}$,we have $\frac{v^2}{r} = k^2rt^2$. Solving for $v$,we get $v^2 = k^2r^2t^2$,which implies $v = krt$. The tangential acceleration $a_t$ is the rate of change of speed: $a_t = \frac{dv}{dt} = \frac{d}{dt}(krt) = kr$. The tangential force $F_t$ responsible for the change in speed is $F_t = m a_t = mkr$. The power $P$ delivered to the particle is given by $P = F_t \cdot v$. Substituting the values,$P = (mkr) \cdot (krt) = mk^2r^2t$.

Class 11 Physics 2 Motion in a Plane Kinematics Circular Motion (Uniform Angular Accelaration)

Difficult

$A$ particle of mass $m$ is moving in a circular path of constant radius $r$ such that its centripetal acceleration $a_c$ is varying with time $t$ as $a_c = k^2rt^2$,where $k$ is a constant. What is the power delivered to the particle by the force acting on it?

A
$2\pi mk^2r^2$
B
$mk^2r^2t$
C
$\frac{mk^4r^2t^5}{3}$
D
Zero

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2 Motion in a Plane

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Kinematics Circular Motion (Uniform Angular Accelaration)

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$A$ particle of mass $m$ is moving in a circular path of constant radius $r$ such that its centripetal acceleration $a_c$ is varying with time $t$ as $a_c = k^2rt^2$,where $k$ is a constant. What is the power delivered to the particle by the force acting on it?

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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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