Question:
A uniform rod of length $\ell$ and mass $m$ is free to rotate in a vertical plane about $A$. The rod initially in horizontal position is released. The initial angular acceleration of the rod is (Moment of inertia of rod about $A$ is $\frac{m \ell^{2}}{3}$ ):
A uniform rod of length $\ell$ and mass $m$ is free to rotate in a vertical plane about $A$. The rod initially in horizontal position is released. The initial angular acceleration of the rod is (Moment of inertia of rod about $A$ is $\frac{m \ell^{2}}{3}$ ):
Updated On: Jul 27, 2022
- $\frac{3 g}{2 \ell}$
- $\frac{2 \ell}{3 g }$
- $\frac{3 g }{2 \ell^{2}}$
- $mg \frac{\ell}{2}$
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Concepts Used:
Electric Field
Electric Field is the electric force experienced by a unit charge.
The electric force is calculated using the coulomb's law, whose formula is:
\(F=k\dfrac{|q_{1}q_{2}|}{r^{2}}\)
While substituting q2 as 1, electric field becomes:
\(E=k\dfrac{|q_{1}|}{r^{2}}\)
SI unit of Electric Field is V/m (Volt per meter).