Question

A small ring of mass m is constrained to slide along a horizontal wire fixed between two rigid supports. The ring is connected to a particle of same mass by an ideal string & the whole system is released from rest as shown in the figure. If the coefficient of friction between ring A and wire is $\frac{3}{5}$ , the ring will start sliding when the connecting string will make an angle $\theta$ with the vertical, then $\theta$ will be (particle is free to move and ring can slide only)

• A. 30^o
• B. 45^o
• C. 60^o
• D. None of these

Answer 1

Answer: (B)

Using energy conservation, we get
$mglcos\theta =\frac{{\text{mv}}^2}{2}$
$\Rightarrow \frac{m{v}^2}{l}=2mgcos\theta$
As the mass is moving in a circular motion, applying circular motion equation,
$\text{T}-\text{mg cos}\theta =2\text{mg cos}\theta$
$\text{T}=3\text{mg cos}\theta$
$Tsin\theta =\mu \left[mg+Tcos\theta \right]$
$3\text{mg cos}\theta sin\theta =\mu \left[\text{mg}+3\text{mg}co{s}^2\theta \right]$
$\theta =45^{{}^{\circ }}$