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Physics - Rotation Question with Solution | TestHub

PhysicsRotationTorque / angular accelerationMedium2 minQB

PhysicsMediumsingle choice

A thin uniform rod of mass M and length $ℓ$ is free to rotate in a vertical plane about a horizontal axis passing through one of its ends. The rod is released from horizontal position as shown in fig. Calculate the shear stress developed at the centre of rod immediately after it is released.

(Given that cross-section area is given as A)

Options:

Answer:

Solution:

Immediately after release: $τ = I α = mg ℓ /2$

$α (\frac{1}{3} m ℓ^{2}) = mg ℓ /2 \Rightarrow α = \frac{3 g}{2 ℓ}$ , $a_{BC} = α \frac{3 L}{4}$

$∴ Fshear + \frac{m}{2} g = \frac{m}{2} \times a_{BC} = \frac{m}{2} \times \frac{3 ℓ}{4} \times \frac{3 g}{2 ℓ}$

shear stress $= \frac{mg}{16 A}$ hence (D).

Stream:JEESubject:PhysicsTopic:RotationSubtopic:Torque / angular acceleration

⏱ 2mℹ️ Source: QB

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