Question

A uniform plank of mass $m=1\text{kg}$ and of sufficient length , which is free to move only in the horizontal direction is placed upon the top of a solid cylinder of mass $2m$ and radius $R$. The plank is attached to a fixed wall by means of a light spring of spring constant $k=7\text{N/m}$. Assuming , there is no slipping between cylinder and the plane system, and between cylinder and the ground. Find the Angular frequency of small oscillations of the system.

• A. $2\text{rad/sec}$
• B. $1\text{rad/sec}$
• C. $3\text{rad/sec}$
• D. $4\text{rad/sec}$

Answer 1

The correct option is A $2\text{rad/sec}$

Suppose that the plank is displaced from its equilibrium position by $x$ at time $t$, the centre of the cylinder is therefore displaced by $\frac{x}{R}$ where $R$ = Radius of the cylinder.
The total energy of the system is given by
$E=K\text{(Plank)}+U\text{(Spring)}+K\text{(Cylinder)}$
$E=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}k{x}^2+\frac{1}{2}{m}_2{v}_2^2+\frac{1}{2}I{\omega }^2$
$E=\frac{1}{2}m{\left(\frac{dx}{dt}\right)}^2+\frac{1}{2}k{x}^2+\frac{1}{2}2m{\left\{\frac{d}{dt}\left(\frac{x}{R}\right)\right\}}^2+\frac{1}{2}(\frac{1}{2}2m.R^2){\left\{\frac{1}{R}\frac{d}{dt}\left(\frac{x}{R}\right)\right\}}^2$
$\Rightarrow E=\frac{1}{2}\left(\frac{7}{4}m\right){\left(\frac{dx}{dt}\right)}^2+\frac{1}{2}k{x}^2$
Since, all the forces acting on the system are conservative, Total energy of the system remains constant.
Differentiating on both sides with respect to time , we get
$\frac{7}{4}m\frac{dx}{dt}\frac{d^{2}x}{d{t}^2}+kx\frac{dx}{dt}=0$
At time $t$ , $\frac{dx}{dt}\ne 0$
$\therefore \frac{7}{4}m\frac{d^{2}x}{d{t}^2}+kx=0$
$\Rightarrow a+\frac{4k}{7m}x=0$
Where $a$ is acceleration,
Comparing the above equation with $a=-{\omega }^{2}x$ we get ,
$\omega =\sqrt{\frac{4k}{7m}}$
From the data given in the question ,
$\omega =\sqrt{\frac{4\times 7}{7\times 1}}=2\text{rad/sec}$
Thus, option (a) is the correct answer.