(A) Given, $\theta = a t^2$.The angular velocity is $\omega = \frac{d\theta}{dt} = 2at$.The tangential velocity is $v = \omega r = 2atr$. Thus, $\frac

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$\frac{v}{t} \sqrt{1+4 a^2 t^4}$

(A) Given, $\theta = a t^2$.The angular velocity is $\omega = \frac{d\theta}{dt} = 2at$.The tangential velocity is $v = \omega r = 2atr$. Thus, $\frac{v}{t} = 2ar$.The tangential acceleration is $a_t = \frac{dv}{dt} = 2ar$.The radial (centripetal) acceleration is $a_n = \frac{v^2}{r} = \frac{(2atr)^2}{r} = 4a^2t^2r$.The total acceleration is $a_{\text{total}} = \sqrt{a_t^2 + a_n^2} = \sqrt{(2ar)^2 + (4a^2t^2r)^2}$.$a_{\text{total}} = \sqrt{4a^2r^2 + 16a^4t^4r^2} = 2ar \sqrt{1 + 4a^2t^4}$.Since $2ar = \frac{v}{t}$, we have $a_{\text{total}} = \frac{v}{t} \sqrt{1 + 4a^2t^4}$.

Class 11 Physics 2 Motion in a Plane Non-uniform Circular Motion (Centrifugal/Centripetal and tangential Accelaration)

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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).

TS EAMCET 2018 All TS EAMCET

A
$\frac{v}{t} \sqrt{1+4 a^2 t^4}$
B
$\frac{v}{t}$
C
$\frac{v}{t}\left(1+4 a^2 t^4\right)$
D
$\sqrt{\left(1+4 a^2 t^4\right)}$

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2 Motion in a Plane

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

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An object is moving on a circular track of radius $450\, m$. At some instant,the object is moving at $30\, m/s$ and gaining speed at a uniform rate of $2\, m/s^2$. Its acceleration at this instant is nearly .......... $m/s^2$.

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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 tangential acceleration varies with time as $a_t = K^2rt^2$ ($K$ is a constant). Select the correct statement.

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