Mass on a semicircular block

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.


(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 …

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Sliding on a block with a circular cut

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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  1. The position
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Circular orbit in a harmonic potential

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?

  1. v=\sqrt{\frac{k}{2 m}} R
  2. v=\sqrt{\frac{k}{m}} R
  3. L=\sqrt{m k} R^{2}
  4. L=\sqrt{\frac{m k}{2}} R^{2}


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 …

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Spherical gas cloud

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]

  1. \frac{K}{2 \pi r^2 m^2 G}
  2. \frac{K}{\pi r^2 m^2 G}
  3. \frac{3 K}{\pi r^2 m^2 G}
  4. \frac{K}{6 \pi r^2 m^2 G}


Consider a small parcel of gas in the cloud with mass \Delta m at a distance r from the center of the cloud (see figure). …

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