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Byju's Answer

Standard XII

Physics

Newton's Second Law

In the arrang...

Question

In the arrangement shown, the rod is freely pivoted at point O and is in contact with the equilateral triangular block which can move on the horizontal frictionless ground. As the block is given a speed $v$ in the forward direction, the rod rotates about point O. Find the angular velocity of the rod in rad/s at the instant when $θ = 30^{\circ}$ . [Take $v = 20 m/s$ , $a = 1 m$ ]

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Solution

$tan θ = \frac{1}{x}$
${sec}^{2} θ \frac{d θ}{d t} = \frac{- 1}{x^{2}} \frac{d x}{d t}$
For $θ = 30^{\circ}, x = \sqrt{3}, sec θ = \frac{2}{\sqrt{3}}$
Substituting we get, $\frac{d θ}{d t} = 5 r a d / s$

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tan θ=1x sec2θdθdt=−1x2dxdt For θ=30∘,x=√3,secθ=2√3 Substituting we get, dθdt=5 rad/s

$tan θ = \frac{1}{x}$
${sec}^{2} θ \frac{d θ}{d t} = \frac{- 1}{x^{2}} \frac{d x}{d t}$
For $θ = 30^{\circ}, x = \sqrt{3}, sec θ = \frac{2}{\sqrt{3}}$
Substituting we get, $\frac{d θ}{d t} = 5 r a d / s$