(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 3-2.Motion in Plane Non-uniform Circular Motion (Centrifugal/Centripetal and tangential Accelaration)

Medium

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
$\frac{v}{t} \sqrt{1+4 a^2 t^4}$
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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3-2.Motion in Plane

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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).

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$A$ stone is tied to one end of a string $50\, cm$ long and is whirled in a horizontal circle with a constant speed. If the stone makes $10$ revolutions in $20\, s$,what is the magnitude of the acceleration of the stone in $cm/s^2$?

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