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, $BD=OP=r-AP$
$=\frac{Mg\cos \theta }{k}-d\cos \theta =\left(\frac{Mg}{k}-d\right)\cos \theta$