A thin rod of length $L$ can rotate in vertical plane about horizontal axis passing through one end of rod without friction. The rod is released when it is horizontal. The initial angular acceleration of rod is:

\frac{2 g}{L}

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\frac{3 g}{L}

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\frac{3 g}{2 L}

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\frac{2 g}{3 L}

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Solution

The correct option is D $\frac{3 g}{2 L}$
Initial torque $τ = M g \frac{L}{2}$
$I = \frac{M L^{2}}{3}$
$I α = τ$
$∴ α \frac{M L^{2}}{3} = \frac{M g L}{2}$
$α = \frac{3 g}{2 L}$

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A thin rod of length L can rotate in vertical plane about horizontal axis passing through one end of rod without friction. The rod is released when it is horizontal. The initial angular acceleration of rod is:

The correct option is D 3g2LInitial torque τ=MgL2I=ML23Iα=τ∴αML23=MgL2α=3g2L

A thin rod of length $L$ can rotate in vertical plane about horizontal axis passing through one end of rod without friction. The rod is released when it is horizontal. The initial angular acceleration of rod is:

The correct option is D $\frac{3 g}{2 L}$
Initial torque $τ = M g \frac{L}{2}$
$I = \frac{M L^{2}}{3}$
$I α = τ$
$∴ α \frac{M L^{2}}{3} = \frac{M g L}{2}$
$α = \frac{3 g}{2 L}$