A uniform circular disc is rolling down an inclined plane of inclination 30° without slipping. Its linear acceleration along the inclined plane is

Solution:

Let $M$ be the mass of the disc and $R$ be its radius. The angle of inclination is $\theta = 30^\circ$.

For a uniform circular disc, the moment of inertia about its center of mass is $I = \frac{1}{2}MR^2$.

When the disc rolls down an inclined plane without slipping, the linear acceleration ($a$) is related to the angular acceleration ($\alpha$) by $a = R\alpha$.

Consider the forces acting on the disc along the inclined plane:

1. Component of gravitational force along the incline: $Mg \sin\theta$ (downwards)

2. Frictional force: $f$ (upwards, opposing motion)

Applying Newton's second law for linear motion:

$Mg \sin\theta - f = Ma$ (Equation 1)

Now, consider the rotational motion about the center of mass. The frictional force $f$ provides the torque:

$\tau = fR$

Also, by Newton's second law for rotation, $\tau = I\alpha$.

So, $fR = I\alpha$

Substitute $I = \frac{1}{2}MR^2$ and $\alpha = \frac{a}{R}$:

$fR = \left(\frac{1}{2}MR^2\right)\left(\frac{a}{R}\right)$

$fR = \frac{1}{2}MRa$

$f = \frac{1}{2}Ma$ (Equation 2)

Substitute Equation 2 into Equation 1:

$Mg \sin\theta - \frac{1}{2}Ma = Ma$

$Mg \sin\theta = Ma + \frac{1}{2}Ma$

$Mg \sin\theta = \frac{3}{2}Ma$

Divide by $M$:

$g \sin\theta = \frac{3}{2}a$

$a = \frac{2}{3}g \sin\theta$

Given $\theta = 30^\circ$, so $\sin 30^\circ = \frac{1}{2}$.

$a = \frac{2}{3}g \left(\frac{1}{2}\right)$

$a = \frac{g}{3}$

Thus, the linear acceleration of the disc along the inclined plane is $\frac{g}{3}$.