A small bead slides on a smooth vertical circular wire of radius 25 m . It is released from rest at a point A on the wire and reaches the lowest point B with a speed of 10 m/s . The angle that the radius vector to point A makes with the downward vertical is: (Take g = 10 m/s ^2 )

Step-by-step solution

Let the vertical height dropped by the bead from point A to the lowest point B be h . Applying the Work-Energy Theorem (or conservation of mechanical energy) between A and B : K + U = 0 1 2 mv^2 - 0 = mgh v^2 = 2gh Substituting the given values ( v = 10 m/s , g = 10 m/s ^2 ): (10)^2 = 2(10)h 100 = 20h h = 5 m From the geometry of the vertical circle, the height h is related to the angle with the downward vertical by: h = R(1 - ) Substituting R = 25 m and h = 5 m : 5 = 25(1 - ) 1 - = 5 25 = 0.2 = 0.8 = 4 5 Therefore