Discussion Forum

A particle of mass m is moving along a circle of radius R such that its tangential acceleration  a_t varies with distance covered x as

a_t =ax² where $\alpha$   is a constant. The kinetic energy,K of the particle varies with the distance as K= $\beta x^{c}$, where $\beta$  and c are constants. The values of $\beta$ and c are

Answer:

Explanation:

 Given, mass of a particel =m

 tangential acceleration of particel ,  $a_{t}=ax^{2} $ and kinetic energy , K=$\beta x^{c}$

Here, as we know that torque,  $\tau=F.R=I\alpha$

$\Rightarrow$     $F.R=I\frac{a_{1}}{R}$              $\left(\because \alpha=\frac{a}{R}\right)$

$\Rightarrow$      $F.R= mR^{2}\frac{aX^{2}}{R}$       ($\because   I= MR^{2}$)

 $\Rightarrow$    $F=max^{2}$  ............(i)

 Hence, the work done is displacement  of dx,

$W=\int f.dx=\int max^{2} dx$

 W=$\frac{1}{3}max^{3}$  ......(iii)

 from work -energy theorm,

  $W= \triangle  KE$

 here, W= work done and $\triangle KE$ = change in kinetic energy,

 $\frac{max^{3}}{3}=\beta x^{2} $     $ (\because K=\beta x^{2})$

 Now, from the above equation , we get

 $\beta =\frac{ma}{3}$ and c=3

 Hence , the correct option is (a)