A particle is suspended vertically from a point O by an inextensible massless string of length L. A vertical line AB is at a distance of L/8 from O as shown. The object is given a horizontal velocity u. At some point, its motion ceases to be circular and eventually the object passes through the line AB. At the instant of crossing AB, its velocity is horizontal. Then u = √[(L/2)(x + 3√3)g]. The value of x is ...
Physics · Mixed Chapter
JEE Mains

Question

A particle is suspended vertically from a point O by an inextensible massless string of length L. A vertical line AB is at a distance of L/8 from O as shown. The object is given a horizontal velocity u. At some point, its motion ceases to be circular and eventually the object passes through the line AB. At the instant of crossing AB, its velocity is horizontal. Then u = √[(L/2)(x + 3√3)g]. The value of x is ...

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: The mass rises from the bottom of the pendulum until tension first becomes zero, at which point it departs from its circular path and becomes a projectile. Step 2: Use energy conservation along the string and the radial condition T=0 to relate the release angle, speed and initial horizontal speed u. For the projectile segment, horizontal acceleration is zero. Step 3: Impose the condition that vertical velocity is zero exactly on the line AB at horizontal distance L/8 from the pivot. The…