Two particles A and B each of mass m are connected by a massless string. A is placed on the rough table. The string passes over a small, smooth peg. B is left from a position making an 
with the vertical. If the minimum coefficient of friction between A and the table is µ min = 3 – N cos 
so that A does not slip during the motion of mass B. Then calculate the value of
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(N = 2)
Sol. Block B rotate in vertical plane. Tension is maximum in string at lowest position. When
block B at lowest position and block A does not slide that means block A not slide at any
position of B.
At lowest position
T – mg = 
⇒ T = mg + 
....
From energy conservation
mg λ (1 – cos θ ) = 
mv 2 ...
from equation and
T = mg + 2mg (1 – cos θ )
= 3mg – 2mg cos θ
for no slipping.
T = μ mg = 3mg – 2mg cos θ
μ min = 3 – 2 cos θ Ans.
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems