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Engineering Mechanics

D Alembert's Principle

A rigid trian...

Question

A rigid triangular body, PQR, with sides of equal length of 1 unit moves on a flat plate. At the instant shown, edge QR is parallel to the x-axis and the body moves such that velocities of points P and R are $V_{P}$ and $V_{R}$ in the x and y directions, respectively. The magnitude of the angular velocity of the body is

2 V_{P}

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\frac{V_{R}}{\sqrt{3}}

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\frac{V_{P}}{\sqrt{3}}

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2 V_{R}

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Solution

The correct option is D $2 V_{R}$

At the given instant, instantaneous Centre

$(I_{0})$ at mid of Q and R, that is S. Let's take $ω$ to be angular velocity of body

Then, $V_{P} = (P S) ω$

$\Rightarrow ω = \frac{V_{P}}{P S} = \frac{V_{P}}{\sqrt{\frac{3}{2}}} = \frac{2 V_{P}}{\sqrt{3}} [∴ P Q = P R = Q R = 1]$ Also, $V_{R} = S R . ω$

$\Rightarrow ω = \frac{V_{R}}{S R} = \frac{V_{R}}{1 / 2} = 2 V_{R}$

Observing options, $2 V_{R}$ is correct i.e. option.

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The correct option is D 2VR At the given instant, instantaneous Centre (I0) at mid of Q and R, that is S. Let's take ω to be angular velocity of body Then, VP=(PS)ω ⇒ω=VPPS=VP√32=2VP√3[∴PQ=PR=QR=1] Also, VR=SR.ω ⇒ω=VRSR=VR1/2=2VR Observing options, 2VR is correct i.e. option.