A peg is placed directly below the pendulum’s point of support at a distance l = 0.5 m from the lowest point of the pendulum swing as shown in the fig. The minimum height h from which the pendulum can be released such that the string remains taut for at least one full revolution around the peg is

Question

A peg is placed directly below the pendulum’s point of support at a distance l = 0.5 m from the lowest point of the pendulum swing as shown in the fig. The minimum height h from which the pendulum can be released such that the string remains taut for at least one full revolution around the peg is <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/9052029efb554ba7e5e5fdb541ac2c51-.png" /> (A) 0.75 m (B) 0.5 m (C) 1.25 m (D) 0.2 m

Tardigrade Admin · Accepted Answer

Critical velocity at highest point =√lg conservation of energy at the starting point and at the highest point implies ∴ m g h =m g 2 l+(1/2) m v2 =m g 2 l+(1/2) m l g=m g l h =(5/2) l=1.25 m

Q. A peg is placed directly below the pendulum’s point of support at a distance l=0.5m from the lowest point of the pendulum swing as shown in the fig. The minimum height h from which the pendulum can be released such that the string remains taut for at least one full revolution around the peg is

0.75 m

0.5 m

1.25 m

0.2 m

Critical velocity at highest point =lg​ conservation of energy at the starting point and at the highest point implies ∴mgh=mg2l+21​mv2 =mg2l+21​mlg=mgl h=25​l=1.25m

Critical velocity at highest point $= l g$ conservation of energy at the starting point and at the highest point implies
$∴ m g h = m g 2 l + \frac{1}{2} m v^{2}$
$= m g 2 l + \frac{1}{2} m l g = m g l$
$h = \frac{5}{2} l = 1.25 m$