A ring of mass ' m ' can slide along a fixed rough vertical rod as shown in fig. The ring is
connected by a spring of spring constant k = 
where 2 R is the natural length of spring. The other end of spring is fixed to the ground at point A at a horizontal distance of 2 R from the base of the rod. If the ring is released from a height of 3R/2 & it reaches the ground with a speed 
, find co-efficient of friction between the rod & ring.
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
Sol. Using work energy theorem,
W f + 
= 
m 
⇒ W f = 
mgR
⇒ f = μ N (as kinetic friction)
W f = 
f dx = 
μ F s cos θ d x
(x = 2 R tan θ ; dx = 2 R sec 2 θ d θ )
F s = k 2 R (sec θ − 1)
W f = μ 
k 2 R (sec θ − 1) cos θ × 2 R sec 2 θ d θ
W f = 4 R 2 μ k 
(sec 2 θ − sec θ ) d , θ 0 = tan − 1 (3/4)
= 4 R 2 μ k 
= –4R 2 μ k [tan θ 0 − ln (sec θ 0 + tan θ 0 )] = 
⇒ 4 R 2 μ k 
= R 2 μ k [3 − 4 ln 2] = 
μ = 
= 
Ans.
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