Question

A disk of radius $R$ and mass $M$ is at equilibrium at position $D$ on the smooth inclined plane which makes an angle $\theta$ with the vertical as shown. The disk's centre is attracted to a point A located at a vertical distance d above the surface as shown. Assume that the force of attraction is proportional to the distance from the disk's centre of mass to point $A$, i.e. assume that $F =- kr$, where $r$ is the distance from the point A to the disk's centre of mass and $k$ is constant. Then, distance $BD$ is

• A. $\left(\frac{Mg}{k} - d\right) \cos\theta$
• B. $\left(\frac{Mg}{k} - d\right) \sin \theta$
• C. $\left(\frac{Mg}{k} + d\right) \cos\theta$
• D. $\left(\frac{Mg}{k} - d\right) \tan \theta$

Answer 1

Answer: (A)

In $\Delta ABP , AP = d \cos \theta$
For equilibrium, $| F | = Mg \cos \theta$
$kr = Mg \cos \theta$
$r =\frac{ Mg \cos \theta}{ k }$
From the figure, $BD = OP = r - AP$
$=\frac{M g \cos \theta}{k}-d \cos \theta=\left(\frac{M g}{k}-d\right) \cos \theta$