Two spherical objects each of radii R and mas | vedclass

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(B) The system is in equilibrium. Let the distance of the center of mass of $m_1$ from the vertical line passing through the suspension point be $r_1$

Vedclass · Accepted Answer

$\frac{2 m_1 R}{\left(m_1+m_2\right) L}$

Vedclass · Answer

(B) The system is in equilibrium. Let the distance of the center of mass of $m_1$ from the vertical line passing through the suspension point be $r_1$ and that of $m_2$ be $r_2$.Since the spheres are in contact,$r_1 + r_2 = 2R$.For rotational equilibrium about the point of contact,the torques due to the weights of the two spheres about the vertical axis passing through the point of contact must balance.$m_1 g r_1 = m_2 g r_2$Substituting $r_1 = 2R - r_2$ into the equation:$m_1 (2R - r_2) = m_2 r_2$$2 m_1 R - m_1 r_2 = m_2 r_2$$2 m_1 R = r_2 (m_1 + m_2)$$r_2 = \frac{2 m_1 R}{m_1 + m_2}$For small angles $	heta$,$\sin 	heta \approx 	heta = \frac{r_2}{L}$.Substituting the value of $r_2$:$	heta = \frac{2 m_1 R}{(m_1 + m_2) L}$

Two spherical objects each of radii $R$ and masses $m_1$ and $m_2$ are suspended using two strings of equal length $L$ as shown in the figure $(R << L)$. The angle $\theta$ which mass $m_2$ makes with the vertical is approximately

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