Q127 / 160 · TS EAMCET 2018 (05 May Shift 1) · Physics

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)

  • Correct answer
    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
    $\sqrt{\left(1+4 a^2 t^4\right)}$

Solution

Given, $\theta=a t^2$ So, angular velocity, $\omega=\frac{d \theta}{d t}=2 a t$ Linear tangential velocity, $$ v=\omega r \quad=2 a t r \quad \frac{v}{t}=2 a r $$ So, tangential linear acceleration of particle is $$ a_t=\frac{d v}{d t}=2 a r $$ and angular acceleration of particle is $$ \alpha=\frac{d \omega}{d t}=2 a $$ Also, normal or radial acceleration of particle is $$ a_n=\frac{v^2}{r}=\frac{4 a^2 t^2 r^2}{r}=4 a^2 t^2 r $$ Total acceleration of particle is $$ \begin{aligned} a_{\text {total }} & =\sqrt{a_t^2+a_n^2}=\sqrt{4 a^2 r^2+16 a^4 t^4 r^2} \\ & =2 a r \sqrt{1+4 a^2 t^4} \\ & =\frac{v}{t} \sqrt{1+4 a^2 t^4} \end{aligned} $$

Hence, the correct answer is option: 'A'.

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