MCQ
$A$ solid uniform disk of mass $m$ rolls without slipping down a fixed inclined plane with an acceleration $a$. The frictional force on the disk due to surface of the plane is :
  • A
    $2\, ma$
  • B
    $3/2\, ma$
  • C
    $ma$
  • $1/2\, ma$

Answer

Correct option: D.
$1/2\, ma$
d
Here as the sphere is rolling down an inclined plane it will produce a torque $\tau=I \alpha$

where $1=$ moment of inertia and $\alpha=$ angular acceleration

Now the frictional force $f=N \mu_{s}$

Where $\mathrm{N}=$ normal force $=m g \cos \theta$ and $\mu_{s}=$ static friction

We also know moment of inertia of sphere $I=\frac{1}{2} m R^{2}$ 'where Reradius of sphere.

and $\alpha=\frac{\text {acceleration}(a)}{R}$

In the given case torque $\tau$ = frictional force $f \times R$

$f R=I \tau$

$f R=\frac{1}{2} m R^{2} \frac{a}{R}$

$f=\frac{m a}{2}$

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