Question

A solid cylinder rolls down on an inclined plane of inclination 30°.The acceleration of the centre of mass of cylinder as it reaches the bottom is

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

Angle of inclination :-

${ \theta = 30 \degree}$

$\huge{\bold{\underline{Explanation:-}}}$

Using the formula:-

$\large{\bold{a_c = \frac{g \sin \theta}{1 + \frac{K^2}{R^2}}}}$

Here K = Radius of gyration.

and, R = Radius of cylinder.

Moment of Inertia of Solid cylinder is

$\large{\bold{I = \frac{1}{2}MR^2}}$

So, the Square of the ratio of radius of gyration to radius of cylinder is given as:-

$\large{\boxed{ \frac{K^2}{R^2} = \frac{1}{2}}}$

Now, Substituting in the acceleration formula.

$\large{\bold{a_c = \frac{g \sin \theta}{1 + \frac{K^2}{R^2}}}}$

$\large{a_c = \frac{10 \sin 30\degree}{1 + \frac{K^2}{R^2}}}$

$\large{a_c = \frac{10 \times 0.5}{1 + \frac{1}{2}}}$

As $\large{ \sin 30 \degree = \frac{1}{2} = 0.5}$

Now,

$\large{a_c = \frac{5}{ \frac{3}{2}}}$

$\large{a_c = \frac{5 \times 2}{3}}$

$\large{a_c = \frac{10}{3}}$

$\huge{\boxed{\boxed{a_c = 3.33 \: m/s^2}}}$

So, the acceleration of centre of mass is 3.33 m/s².