A particle of mass m starts from rest at a di | vedclass

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

Question

(D) The gravitational potential $V$ at a distance $x$ from the centre of a ring of radius $R$ and mass $M$ is given by $V(x) = -\frac{GM}{\sqrt{x^2 +

Vedclass · Accepted Answer

$\sqrt {2\left( {1 - \frac{1}{{\sqrt 2 }}} \right)\frac{{GM}}{R}} $

Vedclass · Answer

(D) The gravitational potential $V$ at a distance $x$ from the centre of a ring of radius $R$ and mass $M$ is given by $V(x) = -\frac{GM}{\sqrt{x^2 + R^2}}$.Applying the principle of conservation of mechanical energy between the initial position $(x = R)$ and the centre $(x = 0)$:$K_i + U_i = K_f + U_f$Since the particle starts from rest,$K_i = 0$.$0 + m V(R) = \frac{1}{2}mv^2 + m V(0)$$0 + m \left( -\frac{GM}{\sqrt{R^2 + R^2}} \right) = \frac{1}{2}mv^2 + m \left( -\frac{GM}{R} \right)$$-\frac{GM}{\sqrt{2}R} = \frac{1}{2}v^2 - \frac{GM}{R}$$\frac{1}{2}v^2 = \frac{GM}{R} - \frac{GM}{\sqrt{2}R} = \frac{GM}{R} \left( 1 - \frac{1}{\sqrt{2}} \right)$$v^2 = \frac{2GM}{R} \left( 1 - \frac{1}{\sqrt{2}} \right)$$v = \sqrt{2\left( 1 - \frac{1}{\sqrt{2}} \right) \frac{GM}{R}}$

$A$ particle of mass $m$ starts from rest at a distance $R$ from the centre and along the axis of a fixed ring of radius $R$ and mass $M$. Its velocity at the centre of the ring is:

Similar Questions

$Assertion$: Space rockets are usually launched in the equatorial line from west to east.
$Reason$: The acceleration due to gravity is minimum at the equator.

Identify the correct option.

For a satellite to orbit around the earth,which of the following must be true?

If a person with a spring balance and a body hanging from it goes up and up in an aeroplane,then the reading of the weight of the body as indicated by the spring balance will

Vedclass Test Series

Exam Paper Generator

Online Exam Module