Discussion Forum

A solid sphere of radius R makes a perfect rolling down on a plane which is inclined to the horizontal axis at an angle $\theta$. If the radius  of gyration is k , then its  acceleration is 

Answer:

Explanation:

 By force balancing  perpendicular to the inclined plane

 N=$mg \cos \theta$

 By forces  balancing  parallel to inclined  plane,

$mg \sin \theta-f_{r}=m\frac{dv}{dt}$

  $\Rightarrow  \frac{dv}{dt}=\frac{mg \sin \theta-f_{r}}{m}\Rightarrow v(t)=\int g \sin \theta dt-\frac{1}{m}\int f_{r}.dt$

 Now, torque,    $\tau= I \frac{d\omega}{dt}= r \times F_{r}$   and    $I=mk^{2}$

$\therefore$      $mk^{2}\frac{d \omega}{dt}=rF_{r}$    or   $\frac{d\omega}{dt}= \frac{rF_{r}}{mk^{2}}$

$\Rightarrow$      $\omega(t)=\frac{R}{mk^{2}}\int f_{r}dt$

or    $\frac{1}{m}\int f_{r}  dt= \frac{k^{2}}{R}\omega(t)=\frac{k^{2}}{R^{2}}v(t)$   (using $\omega$=v/R)

  $\Rightarrow$     $v(t)=\int g \sin \theta dt- \frac{k^{2}}{R^{2}}v(t)$

 or    $v(t)=\frac{1}{\left(1+\frac{k^{2}}{R^{2}}\right)}\int g\sin \theta dt$

 So, acceleration, 

$a(t)=\frac{dv(t)}{dt}=\frac{g \sin \theta}{\left(1+\frac{k^{2}}{R^{2}}\right)}$