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Newton's Second Law: Rotatory Version

An L-shaped o...

Question

An L-shaped object, made of thin rods of uniform mass density, is suspended with a string as showm=n in figure. If $A B = B C$ , and the angle made by $A B = B$ , and the angle made by $A B$ with downward vertical is $θ$ , then:

tan θ = \frac{2}{\sqrt{3}}

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tan θ = \frac{1}{3}

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tan θ = \frac{1}{2}

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tan θ = \frac{1}{2 \sqrt{3}}

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Solution

The correct option is A $tan θ = \frac{1}{3}$
Let the mass of one of is $m$ .

Balancing torque about the point

$m g (C_{1} P) = m g (C_{2} N)$

$m g (\frac{L}{2} sin θ) = m g (\frac{L}{2} cos θ - L sin θ)$

$\Rightarrow \frac{3}{2} m g L sin θ = \frac{m g L}{2} cos θ$

$\Rightarrow tan θ = \frac{1}{3}$

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An L-shaped object, made of thin rods of uniform mass density, is suspended with a string as showm=n in figure. If AB=BC, and the angle made by AB=B, and the angle made by AB with downward vertical is θ, then:

The correct option is A tanθ=13Let the mass of one of is m.Balancing torque about the point mg(C1P)=mg(C2N) mg(L2sinθ)=mg(L2cosθ−Lsinθ)⇒32mgLsinθ=mgL2cosθ⇒tanθ=13

An L-shaped object, made of thin rods of uniform mass density, is suspended with a string as showm=n in figure. If $A B = B C$ , and the angle made by $A B = B$ , and the angle made by $A B$ with downward vertical is $θ$ , then: