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Standard XII

Physics

Newton's Second Law: Rotatory Version

A uniform tri...

Question

A uniform triangular plate ABC of moment of mass m and inertia I (about an axis passing through A and perpendicular to plane of the plate) can rotate freely in the vertical plane about point 'A' as shown in figure. The plate is released from the position shown in the figure. Line AB is horizontal. The acceleration of center of mass just after the release of plate is :

\frac{m g a^{2}}{\sqrt{3} I}

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\frac{m g a^{2}}{4 I}

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\frac{m g a^{2}}{2 \sqrt{3} I}

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\frac{m g a^{2}}{3 I}

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Solution

The correct option is A $\frac{m g a^{2}}{2 \sqrt{3} I}$
Torque about: $m g \frac{a}{2} = I α$
$\Rightarrow α = \frac{m g a}{2 I}$
$\Rightarrow A c c e l e r a t i o n = \frac{a}{\sqrt{3}} α = \frac{m g a^{2}}{2 \sqrt{3} I}$

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The correct option is A mga22√3ITorque about: mga2=Iα⇒α=mga2I⇒Acceleration=a√3α=mga22√3I

The correct option is A $\frac{m g a^{2}}{2 \sqrt{3} I}$
Torque about: $m g \frac{a}{2} = I α$
$\Rightarrow α = \frac{m g a}{2 I}$
$\Rightarrow A c c e l e r a t i o n = \frac{a}{\sqrt{3}} α = \frac{m g a^{2}}{2 \sqrt{3} I}$