A frictionless wire is fixed between A and B inside a ring of radius R. A bead slips along the wire. The time taken by the bead to slip from A to B will be
⇒AB=0+12gcosθt2
or,2Rcosθ=12gcosθt2
∴t=√4Rg=2√Rg
It gives required time for bead to slip from A to B.
Why this question? Tip: Bead is constrained to move along AB, hence acceleration along AB is provided by the component of weight along AB, i.e. mgcosθ. |