The moment of inertia of a solid body of mass m relative to the O-axis is equal to J. The center of mass of the body is located at a distance R from this axis. Find the force acting on the axis when a force F is briefly applied to a solid body perpendicular to the segment of length x that connects the point of application of the force and the axis. At which x is the smallest force acting on the axis?

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    <h3> Statement </h3>
    <p>
        $2.7.44.$ The moment of inertia of a solid body of mass $m$ relative to the $O$-axis is equal to $J$. The center of mass of the body is located at a distance $R$ from this axis. Find the force acting on the axis when a force $F$ is briefly applied to a solid body perpendicular to the segment of length $x$ that connects the point of application of the force and the axis. At which $x$ is the smallest force acting on the axis?

    <h3>Solution</h3>
    <p>

Force $\vec{F}$ creates an angular momentum $\vec{F}\cdot x$ (with $x$ perpendicular to $\vec{F}$): $$\tau_0=F\cdot x$$ $$\tau_f=J\cdot\varepsilon$$ In steady-state mode, the angular moments will be equal $$\tau_0=\tau_f=Fx=J\varepsilon$$ From where $$\varepsilon = \frac{Fx}{J}$$ Newton's second law $$F-F_0=m\varepsilon R$$ $$F_0=F-mR\cdot \frac{Fx}{J}\Leftrightarrow \boxed{F_0=F\left(1-\frac{mRx}{J}\right)}$$ From here, the minimum values of $F_0 = 0$, described by the expression $$F_0=0\Leftrightarrow 1-\frac{mRx}{J}=0$$ which will be observed at length that connects the point of application of the force and the axis $$\boxed{x=\frac{J}{mR}}$$

    <h4>Answer</h4>
    <p>
        $F'=F(1-mRx/J)$. With $x=J/(mR),\~F'=0$
    </p>
    <p style="text-align: right; font-style: italic; font-size: 14;">
      Almaskhan Arsen<br>
    </p>


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