Point masses m_1 and m_2 are placed at the op | vedclass

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(A) The moment of inertia $I$ of the system about an axis passing through point $P$ at a distance $x$ from $m_1$ is given by:$I = m_1 x^2 + m_2 (L - x

Vedclass · Accepted Answer

$x = \frac{m_2 L}{m_1 + m_2}$

Vedclass · Answer

(A) The moment of inertia $I$ of the system about an axis passing through point $P$ at a distance $x$ from $m_1$ is given by:$I = m_1 x^2 + m_2 (L - x)^2$According to the work-energy theorem,the work done $W$ to set the rod rotating with angular velocity $\omega_0$ is equal to the rotational kinetic energy:$W = \frac{1}{2} I \omega_0^2 = \frac{1}{2} [m_1 x^2 + m_2 (L - x)^2] \omega_0^2$For the work $W$ to be minimum,we set the derivative with respect to $x$ to zero:$\frac{dW}{dx} = \frac{1}{2} \omega_0^2 [2 m_1 x + 2 m_2 (L - x)(-1)] = 0$$m_1 x - m_2 (L - x) = 0$$m_1 x - m_2 L + m_2 x = 0$$(m_1 + m_2) x = m_2 L$$x = \frac{m_2 L}{m_1 + m_2}$

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