A body slides down on a frictionless track which ends in a circular loop of diameter D . The minimum height h in terms of D so that the body may just complete the circular loop, is

Question

A body slides down on a frictionless track which ends in a circular loop of diameter D . The minimum height h in terms of D so that the body may just complete the circular loop, is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-mpztlpwr2ah8kdyw.jpg" /> (A) h=(5/2)D (B) h=(3/2)D (C) h=(5/4)D (D) h=2D

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-mpztlpwr2ah8lqtr.jpg" /> By conservation of energy, (1/2) m v2 = m g h v = √2 g h ...(i) For looping the loop, the lower velocity must be greater than √5 g r vmin = √5 g r = √(5 g D/2) ...(ii) From equation (i) and (ii), 2 g h = (5 g D/2) h = (5 D/4)

Q. A body slides down on a frictionless track which ends in a circular loop of diameter $D$ . The minimum height $h$ in terms of $D$ so that the body may just complete the circular loop, is

$h = \frac{5}{2} D$

$h = \frac{3}{2} D$

$h = \frac{5}{4} D$

$h = 2 D$

By conservation of energy,
$\frac{1}{2} m v^{2} = m g h$
$v = 2 g h$ ...(i)
For looping the loop, the lower velocity must be greater than $5 g r$
$v_{min} = 5 g r = \frac{5 g D}{2}$ ...(ii)
From equation (i) and (ii),
$2 g h = \frac{5 g D}{2}$
$h = \frac{5 D}{4}$

Q. A body slides down on a frictionless track which ends in a circular loop of diameter D . The minimum height h in terms of D so that the body may just complete the circular loop, is

h=25​D

h=23​D

h=45​D

h=2D