A small body slides down a smooth uneven surf | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the mass of the body be $m$. At point $A$,the normal force $N$ provides the centripetal force required for circular motion. Since the weight $

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the mass of the body be $m$. At point $A$,the normal force $N$ provides the centripetal force required for circular motion. Since the weight $mg$ acts vertically downwards and the normal force acts horizontally towards the center $O$,the net force on the body at $A$ is the vector sum of $N$ and $mg$.The magnitude of the net force is $F_A = \sqrt{N^2 + (mg)^2}$.Given $F_A = \sqrt{2} mg$,we have $\sqrt{N^2 + (mg)^2} = \sqrt{2} mg$.Squaring both sides,$N^2 + m^2g^2 = 2m^2g^2$,which gives $N^2 = m^2g^2$,so $N = mg$.The centripetal force equation at $A$ is $N = \frac{mv_A^2}{R}$,so $mg = \frac{mv_A^2}{R}$,which implies $v_A^2 = Rg$.By the law of conservation of energy,the total energy at the starting point $P$ (height $H$) equals the total energy at point $A$ (height $R$):$mgH = \frac{1}{2}mv_A^2 + mgR$.Substituting $v_A^2 = Rg$:$mgH = \frac{1}{2}m(Rg) + mgR = \frac{3}{2}mgR$.Therefore,$H = \frac{3}{2}R$.

$A$ 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 $\sqrt{2}$ times its weight?

Similar Questions

$A$ particle is moving along a horizontal circle of radius $r$ under a centripetal force $F = -\frac{c}{r^2}$,where $c$ is a constant. Then,the total energy of the particle is

$A$ body attached to one end of a string performs motion along a vertical circle. Its centripetal acceleration,when the string is horizontal,will be [ $g$ = acceleration due to gravity]

$A$ body of mass $m \ kg$ slides from rest along the curve of a vertical circle from point $A$ to $B$ on a frictionless path. The velocity of the body at $B$ is: (Given: $R = 14 \ m$,$g = 10 \ m/s^2$,and $\sqrt{2} = 1.4$) (in $m/s$)

$A$ particle originally at rest at the highest point of a smooth vertical circle is slightly displaced. It will leave the circle at a vertical distance $h$ below the highest point such that

$A$ bucket filled with water is revolved in a vertical circle of radius $4 \ m$ and the water just does not fall down. The time period of revolution will be ........ $s$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module