A hemispherical shell of mass 2M and radius 6 | vedclass

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(D) The center of mass of a hemispherical shell of radius $6R$ is at a distance of $3R$ from its center. Let the mass of the shell be $m_1 = 2M$ and t

Vedclass · Accepted Answer

$\omega_2 = \sqrt{\frac{GM}{72R^3}}$

Vedclass · Answer

(D) The center of mass of a hemispherical shell of radius $6R$ is at a distance of $3R$ from its center. Let the mass of the shell be $m_1 = 2M$ and the point mass be $m_2 = M$. The distance between them is $d = 3R$.For a two-body system,the radii of the circular orbits about the center of mass are $r_1 = \frac{m_2 d}{m_1 + m_2} = \frac{M(3R)}{2M + M} = R$ and $r_2 = \frac{m_1 d}{m_1 + m_2} = \frac{2M(3R)}{2M + M} = 2R$.The gravitational force between them is $F = \frac{G m_1 m_2}{d^2} = \frac{G(2M)(M)}{(3R)^2} = \frac{2GM^2}{9R^2}$.For the point mass $M$,the centripetal force is provided by the gravitational force: $M \omega^2 r_2 = F$.$M \omega^2 (2R) = \frac{2GM^2}{9R^2}$.$\omega^2 = \frac{GM}{9R^3}$.Since both bodies rotate with the same angular speed $\omega_1 = \omega_2 = \omega$,we have $\omega = \sqrt{\frac{GM}{9R^3}}$.Looking at the options,there seems to be a discrepancy in the provided options based on the standard calculation. However,if we re-evaluate the force calculation assuming the distance between the center of mass of the shell and the point mass is $3R$,the result is $\omega = \sqrt{\frac{GM}{9R^3}}$. Given the options,$D$ is the closest in form,but based on the physics,the correct angular speed is $\sqrt{\frac{GM}{9R^3}}$.

List-$I$	List-$II$
$(A)$ Kinetic energy of planet	$(1)$ $-\frac{GMm}{a}$
$(B)$ Gravitational potential energy of Sun-planet system	$(2)$ $\frac{GMm}{2a}$
$(C)$ Total mechanical energy of planet	$(3)$ $\frac{GM}{r}$
$(D)$ Escape energy at the surface of planet for unit mass object	$(4)$ $-\frac{GMm}{2a}$

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