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
$mgl cos \theta = \frac{\text{mv}^{2}}{2}$
$\Rightarrow \frac{mv^{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 $
$Tsin\theta =\mu \left[mg + Tcos\theta \right]$
$3 \text{mg cos} \theta sin \theta = \mu \left[\text{mg} + 3 \text{mg} cos^{2} \theta \right]$
$\theta = 4 5^{^\circ }$