A small body slides down a smooth uneven surface from a height H, which eventually emerges into a circular loop of radius R(

Question

A small body slides down a smooth uneven surface from a height H, which eventually emerges into a circular loop of radius R(< H) what should be the value of H, so that the force on the body at A is √2 times its weight? <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/305808809d983b72683c81f3fc53c77c-.png" /> (A) H=(3R/2) (B) H=5R (C) H=(5R/2) (D) None of these

Tardigrade Admin · Accepted Answer

Force on body at A=√2 mg ⇒ F=√Fg2+Fc2 ⇒ Fc=m g=m v2 / R ⇒ v2=g R Using energy conservation m g H=m v2 / 2+m g R =3 m g R / 2 H=3 R / 2

Q. A small body slides down a smooth uneven surface from a height $H,$ which eventually emerges into a circular loop of radius $R (< H)$ what should be the value of $H,$ so that the force on the body at A is $2$ times its weight?

$H = \frac{3 R}{2}$

$H = 5 R$

$H = \frac{5 R}{2}$

None of these

Force on body at $A = 2 m g$
$\Rightarrow F = F_{g}^{2} + F_{c}^{2}$
$\Rightarrow F_{c} = m g = m v^{2} / R$
$\Rightarrow v^{2} = g R$
Using energy conservation
$m g H = m v^{2} /2 + m g R$
$= 3 m g R /2$
$H = 3 R /2$

Q. A small body slides down a smooth uneven surface from a height H, which eventually emerges into a circular loop of radius R(<H) what should be the value of H, so that the force on the body at A is 2​ times its weight?

H=23R​

H=5R

H=25R​