A small disc is on top of a hemisphere of radius R . smallest horizontal velocity v that should be g

English

Gujarati

Hindi

5.Work, Energy, Power and Collision

medium

A small disc is on the top of a hemisphere of radius $R$. What is the smallest horizontal velocity v that should be given to the disc for it to leave the hemisphere and not slide down it ? [There is no friction]

A

$v = \sqrt {2gR} $

B

$v = \sqrt {gR} $

C

$v = \frac{g}{R}$

D

$v = \sqrt {{g^2}R} $

Solution

(b) $m g-N=\frac{m v^{2}}{R}$
$N=0 \Rightarrow v=\sqrt{g R}$

Standard 11

Physics

Share

Similar Questions

A stone of mass $1\, kg$ tied to a light inextensible string of length $L = \frac{{10}}{3}m$ is whirling in a circular path of radius $L$ in a vertical plane. If the ratio of the maximum tension in the string to the minimum tension in the string is $4$ and if $g$ is taken to be $10m/{\sec ^2}$, the speed of the stone at the highest point of the circle is ……. $m/\sec$

hard

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

medium

A stone of mass m is tied to a string and is moved in a vertical circle of radius $ r$ making n revolutions per minute. The total tension in the string when the stone is at its lowest point is

medium

$A$ particle of mass $m$ is released from $a$ height $H$ on $a$ smooth curved surface which ends into a vertical loop of radius $R$, as shown Choose the correct alternative $(s)$ if $H = 2R$

hard

$A$ small ball $B$ of mass $m$ is suspended with light inelastic string of length $L$ from $a$ block $A$ of same mass $m$ which can move on smooth horizontal surface as shown in the figure. The ball is displaced by angle $\theta$ from equilibrium position & then released. The displacement of block when ball reaches the equilibrium position is

hard