A rod of mass M and length L is placed on a horizontal plane with one end hinged about the vertical axis- A horizontal force of F-mg3 is applied at a distance L3 from the hinged end- The angular acceleration of the rod will be-

τ = Torque #160; τ = Iα #160; α = τ/I #160;= rf/I = (L/3)(mg/3)/ML^2/3 #160; = mgL/3ML^2 #160; α ...

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Byju's Answer

Standard XII

Physics

The Complete Equilibrium

A rod of mass...

Question

A rod of mass $M$ and length $L$ is placed on a horizontal plane with one end hinged about the vertical axis. A horizontal force of $F = \frac{m g}{3}$ is applied at a distance $\frac{L}{3}$ from the hinged end. The angular acceleration of the rod will be:-

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Solution

$τ$ = Torque

$τ$ = $I α$

$α$ = $\frac{τ}{I}$

= $\frac{r f}{I}$ = $\frac{(L / 3) (m g / 3)}{M L^{2} / 3}$

= $\frac{m g L}{3 M L^{2}}$
$α$ = $\frac{g}{3 L}$

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