Discussion Forum

A particle moves in a circle  with speed v varying with time as v(t)=2t. The total acceleration of particle after it completes 2 rounds of circle 

Answer:

Explanation:

 Instantaneous velocity of particle in circular motion is 

 $v=r\omega=at$  , (let v=at)

Then , $\omega=\frac{d\theta}{dt}=\frac{at}{r}$

So,   $\int_{0}^{2 \pi n}\theta=\int_{0}^{t} \frac{at}{r} dt$

 n= number of rounds=2(given)

$\Rightarrow$   $2\pi n=\frac{at^{2}}{2r}\Rightarrow t^{2}=\frac{4\pi nr}{a}$

Radial acceleration , $a_{r}=\frac{v^{2}}{r} = \frac{a^{2}t^{2}}{r}$

 $\Rightarrow$   $a_{r}=\frac{a^{2}}{r} \times \frac{4 \pi nr}{a}$

 $\Rightarrow$    $a_{r}=4 \pi na$

Tangential acceleration , $a_{t}=\frac{dv}{dt}=a$

 $\therefore$   Total acceleration,

$a=\sqrt{a_{t}^{2}+a_{r}^{2}}=\sqrt{a^{2}+(4\pi na)^{2}}$

$=a\sqrt{1+(4\pi n)^{2}}$

$=2\sqrt{1+64 \pi^{2}}$    [given a=2,n=2]