Q125 / 160 · TS EAMCET 2018 (05 May Shift 1) · Physics
A particle moves in a circle with speed $v$ varying with time as $v(t)=2 t$. The total acceleration of the particle after it completes 2 rounds of cycle is
- A$16 \pi$
- Correct answerB$2 \sqrt{1+64 \pi^2}$
- C$2 \sqrt{1+49 \pi^2}$
- D$14 \pi$
Solution
Instantaneous velocity of particle in circular motion is
$v=r \omega=a t$,
(let $v=a t)$
Then, $\quad \omega=\frac{d \theta}{d t}=\frac{a t}{r}$
So, $\quad \int_0^{2 \pi n}=\int_0^t \frac{a t}{r} d t$,
$n=$ number of rounds $=2$ (given)
$\Rightarrow \quad 2 \pi n=\frac{a t^2}{2 r} \Rightarrow t^2=\frac{4 \pi n r}{a}$
Radial acceleration, $a_r=\frac{v^2}{r}=\frac{a^2 t^2}{r}$
$$
\begin{aligned}
\Rightarrow & a_r & =\frac{a^2}{r} \times \frac{4 \pi n r}{a} \\
\Rightarrow & a_r & =4 \pi n a
\end{aligned}
$$
Tangential acceleration, $a_t=\frac{d v}{d t}=a$
$\therefore$ Total acceleration,
$$
\begin{aligned}
a & =\sqrt{a_t^2+a_r^2}=\sqrt{a^2+(4 \pi n a)^2} \\
& =a \sqrt{1+(4 \pi n)^2} \\
& \left.=2 \sqrt{1+64 \pi^2} \quad \text { [given } a=2, n=2\right]
\end{aligned}
$$
Hence, the correct answer is option: 'B'.