A uniform rod of mass m and length l rotates | vedclass

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(D) Consider an element of the rod of length $dx$ at a distance $x$ from the axis of rotation.The mass of this element is $dm = \frac{m}{l} dx$.The ce

Vedclass · Accepted Answer

$\frac{1}{2} \frac{m \omega^2}{l} \left( l^2 - x^2 \right)$

Vedclass · Answer

(D) Consider an element of the rod of length $dx$ at a distance $x$ from the axis of rotation.The mass of this element is $dm = \frac{m}{l} dx$.The centripetal force required for this element to rotate is $dF = (dm) \omega^2 x = \frac{m}{l} \omega^2 x dx$.This force is provided by the difference in tension across the element,i.e.,$dF = T(x) - T(x+dx) = -dT$.Integrating from $x$ to $l$ (where tension $T=0$ at $x=l$):$\int_{T(x)}^{0} -dT = \int_{x}^{l} \frac{m}{l} \omega^2 x dx$.$T(x) = \frac{m \omega^2}{l} \left[ \frac{x^2}{2} \right]_{x}^{l} = \frac{m \omega^2}{l} \left( \frac{l^2 - x^2}{2} \right)$.Therefore,$T = \frac{1}{2} \frac{m \omega^2}{l} (l^2 - x^2)$.

$A$ uniform rod of mass $m$ and length $l$ rotates in a horizontal plane with an angular velocity $\omega$ about a vertical axis passing through one end. The tension in the rod at a distance $x$ from the axis is

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