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 $\tau=I \alpha$
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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