Physics - Rotation Question with Solution | TestHub

Let the linear acceleration of the disc down the incline be $a$ and its angular acceleration be $\alpha$.

The forces acting on the disc along the inclined plane are the component of gravity downwards, $mg\sin \theta$, and the frictional force $f$ acting upwards along the incline to oppose the angular deceleration.

The equation for linear motion is:
$mg\sin \theta -f=ma$

The equation for rotational motion about the center of mass is:
$fR=I\alpha$

For pure rolling, the relationship between linear and angular acceleration is:
$a=R\alpha$

The moment of inertia of a uniform disc about its central axis is:
$I=\frac{1}{2}m{R}^2$

Substituting $I$ and $\alpha =\frac{a}{R}$ into the torque equation:
$fR=\left(\frac{1}{2}m{R}^2\right)\left(\frac{a}{R}\right)$

$f=\frac{1}{2}ma\Rightarrow a=\frac{2f}{m}$

Substitute this value of $a$ into the linear motion equation:
$mg\sin \theta -f=m\left(\frac{2f}{m}\right)$

$mg\sin \theta -f=2f$

$3f=mg\sin \theta$

$f=\frac{mg\sin \theta }{3}$

Given that $\theta =30^{\circ }$ and $\sin 30^{\circ }=\frac{1}{2}$:
$f=\frac{mg}{3}\times \frac{1}{2}=\frac{mg}{6}$

Answer: $\frac{mg}{6}$