(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 3-2.Motion in 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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3-2.Motion in Plane

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Kinematics Circular Motion (Uniform Angular Accelaration)

In the $P^{\text{th}}$ second,a particle describes an angular displacement of $\beta \text{ rad}$. If it starts from rest,the angular acceleration is:

DifficultMHT CET 2023

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$A$ wheel which is initially at rest is subjected to a constant angular acceleration about its axis. It rotates through an angle of $15^{\circ}$ in time $t \ s$. The increase in angle through which it rotates in the next $2t \ s$ is (in $^{\circ}$)

DifficultAP EAMCET 2014

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$A$ wheel of radius $30 \ cm$ is driven by a belt. Its initial angular speed is $2 \ rev/s$. Starting at this speed,it comes to rest after a length of $25 \ m$ of the belt has passed over the wheel. The angular deceleration of the wheel is ....... $rad \ s^{-2}$.

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The angular speed of a truck wheel is increased from $900 \, rpm$ to $2460 \, rpm$ in $26 \, s$. The number of revolutions made by the truck engine during this time is (assuming the acceleration to be uniform).

MediumJEE MAIN 2021

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$A$ particle moves along a circular path of radius $r = \frac{20}{\pi} \ m$ with a constant tangential acceleration. If the velocity of the particle at the end of the second revolution is $80 \ m/s$,what is the tangential acceleration?

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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?

Similar Questions

In the $P^{\text{th}}$ second,a particle describes an angular displacement of $\beta \text{ rad}$. If it starts from rest,the angular acceleration is:

$A$ wheel which is initially at rest is subjected to a constant angular acceleration about its axis. It rotates through an angle of $15^{\circ}$ in time $t \ s$. The increase in angle through which it rotates in the next $2t \ s$ is (in $^{\circ}$)

$A$ wheel of radius $30 \ cm$ is driven by a belt. Its initial angular speed is $2 \ rev/s$. Starting at this speed,it comes to rest after a length of $25 \ m$ of the belt has passed over the wheel. The angular deceleration of the wheel is ....... $rad \ s^{-2}$.

The angular speed of a truck wheel is increased from $900 \, rpm$ to $2460 \, rpm$ in $26 \, s$. The number of revolutions made by the truck engine during this time is (assuming the acceleration to be uniform).

$A$ particle moves along a circular path of radius $r = \frac{20}{\pi} \ m$ with a constant tangential acceleration. If the velocity of the particle at the end of the second revolution is $80 \ m/s$,what is the tangential acceleration?

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