(B) The centripetal acceleration is given by $a_c = \frac{v^2}{r}$.Given $a_c = k^2rt^2$, we have $\frac{v^2}{r} = k^2rt^2$, which implies $v^2 = k^2r

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(B) The centripetal acceleration is given by $a_c = \frac{v^2}{r}$.Given $a_c = k^2rt^2$, we have $\frac{v^2}{r} = k^2rt^2$, which implies $v^2 = k^2r^2t^2$.Taking the square root, the speed of the particle is $v = krt$.The tangential acceleration is $a_t = \frac{dv}{dt} = \frac{d}{dt}(krt) = kr$.The tangential force is $F_t = ma_t = mkr$.The power delivered to the particle is $P = F_t \cdot v = (mkr) \cdot (krt) = mk^2r^2t$.

Class 11 Physics 3-2.Motion in Plane Non-uniform Circular Motion (Centrifugal/Centripetal and tangential 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 is varying with time as $a_c = k^2rt^2$. The power delivered to the particle by the forces acting on it is $(k = \text{constant})$

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A
$m^2k^2r^2t^2$
B
$mk^2r^2t$
C
$2\pi mk^2r^2t$
D
$mk^4r^2t^3$

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3-2.Motion in Plane

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Non-uniform Circular Motion (Centrifugal/Centripetal and tangential Accelaration)

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In non-uniform circular motion,the ratio of tangential to radial acceleration is ($r$ is the radius of the circle,$v$ is the speed of the particle,$\alpha$ is the angular acceleration).

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$A$ cyclist is riding with a speed of $27 \; km/h$. As he approaches a circular turn on the road of radius $80 \; m$,he applies brakes and reduces his speed at the constant rate of $0.50 \; m/s^2$. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?

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$A$ cyclist is riding with a speed of $36 \,km/h$. As he approaches a circular turn on the road of radius $50 \,m$, he applies brakes and reduces his speed at the constant rate of $0.5 \,m/s^2$. The magnitude and direction of the net acceleration of the cyclist on the circular turn are respectively:

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$A$ particle moves in a circular path of radius $r$ with speed $v.$ It then increases its speed to $2v$ while travelling along the same circular path. The centripetal acceleration of the particle has changed by a factor of

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Consider a wheel rotating around a fixed axis. If the rotation angle $\theta$ varies with time as $\theta=a t^2$, then the total acceleration of a point $A$ on the rim of the wheel is ($v$ being the tangential velocity).

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$A$ particle of mass $m$ is moving in a circular path of constant radius $r$ such that its centripetal acceleration is varying with time as $a_c = k^2rt^2$. The power delivered to the particle by the forces acting on it is $(k = \text{constant})$

Similar Questions

In non-uniform circular motion,the ratio of tangential to radial acceleration is ($r$ is the radius of the circle,$v$ is the speed of the particle,$\alpha$ is the angular acceleration).

$A$ cyclist is riding with a speed of $27 \; km/h$. As he approaches a circular turn on the road of radius $80 \; m$,he applies brakes and reduces his speed at the constant rate of $0.50 \; m/s^2$. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?

$A$ particle moves in a circular path of radius $r$ with speed $v.$ It then increases its speed to $2v$ while travelling along the same circular path. The centripetal acceleration of the particle has changed by a factor of

Consider a wheel rotating around a fixed axis. If the rotation angle $\theta$ varies with time as $\theta=a t^2$, then the total acceleration of a point $A$ on the rim of the wheel is ($v$ being the tangential velocity).

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