Consider a wheel rotating around a fixed axis | vedclass

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

Vedclass · Accepted Answer

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

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(A) Given,$	heta = a t^2$.The angular velocity is $\omega = \frac{d	heta}{dt} = 2at$.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 $a_{	ext{total}} = \sqrt{a_t^2 + a_n^2} = \sqrt{(2ar)^2 + (4a^2t^2r)^2}$.$a_{	ext{total}} = \sqrt{4a^2r^2 + 16a^4t^4r^2} = 2ar \sqrt{1 + 4a^2t^4}$.Since $2ar = \frac{v}{t}$,we have $a_{	ext{total}} = \frac{v}{t} \sqrt{1 + 4a^2t^4}$.

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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