Centrifugal force – JEE First

Mass on a semicircular block

February 1st, 2022jeefirst0 Comments

A heavy particle of mass $m$ is placed at the top of a semicircular block of radius $R$ . Find the height at which the particle falls off, assuming (i) the block is fixed to the ground, and (ii) the block has a mass $M$ and is free to move. Assume all surfaces are frictionless.

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Related problem: Sliding on a block with a circular cut.

Solution:

(i) We first consider the case where the block is fixed to the ground. As the mass slides down the block, there are three forces acting on it: the weight $mg$ , the centrifugal force $m R \dot{\theta}^2$ , and …

Sliding on a block with a circular cut

January 30th, 2022jeefirst0 Comments

JEE Advanced 2017 Paper 1, Question 2

A block of mass $M$ has a circular cut with a frictionless surface as shown. The block rests on the horizontal frictionless surface of a fixed table. Initially the right edge of the block is at $x=0$ , in a co-ordinate system fixed to the table. A point mass $m$ is released from rest at the topmost point of the path as shown and it slides down. When the mass loses contact with the block, its position is $x$ and the velocity is $v$ . At that instant, which of the following options is/are correct?

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The position

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Circular orbit in a harmonic potential

January 21st, 2022jeefirst0 Comments

JEE Advanced 2018 Paper 1, Question 1

The potential energy of a particle of mass $m$ at a distance $r$ from a fixed point $O$ is given by $V(r)=k r^{2} / 2$ , where $k$ is a positive constant of appropriate dimensions. This particle is moving in a circular orbit of radius $R$ about the point $O$ . If $v$ is the speed of the particle and $L$ is the magnitude of its angular momentum about $O$ , which of the following statements is (are) true?

$v=\sqrt{\frac{k}{2 m}} R$
$v=\sqrt{\frac{k}{m}} R$
$L=\sqrt{m k} R^{2}$
$L=\sqrt{\frac{m k}{2}} R^{2}$

Solution

The force due to the given potential is

(1) $\begin{equation*} {\bf F} = -\frac{\partial V}{\partial r} \hat{\bf r} = -k r \hat{\bf r} . \end{equation*}$

The mass also experiences a centrifugal force due to its circular motion,

(2) $\begin{equation*} {\bf F}_{\rm cent} = \frac{m v^2}{r} \hat{\bf r} \end{equation*}$

For the …

Spherical gas cloud

December 30th, 2021jeefirst0 Comments

JEE Advanced 2019 Paper 1, Question 1

Consider a spherical gas cloud of mass density $\rho(r)$ in free space where $r$ is the radial distance from its center. The gaseous cloud is made of particles of equal mass $m$ moving in circular orbits about the common center with the same kinetic energy $K$ . The force acting on the particles is their mutual gravitational force. If $\rho(r)$ is constant in time, the particle number density $n(r) = \rho(r)/m$ is [ $G$ is the universal gravitational constant]