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A cylindrical block of mass m and height h floats vertically in a liquid which is twice as dense as the block, as shown in the figure. A uniform rod of mass m/2 and length I is hinged to axis of the cylinder at A and to the wall of the vessel at B. The rod can rotate freely about point B. It is observed that the rod is horizontal in its equilibrium position. Angular frequency of small vertical oscillations of the cylinder will be m/2 B 12g V 7h

Asked by u_m67 | 09 Apr, 2022, 08:43: AM

Expert Answer

Initially cylinder of mass m and height h floats in the liquid at a depth h/2 . Let it be pressed to an additional depth y and be released .

The total torque acting on the rod is shown in figure

Total torque τ = [ ( F_b - F_g ) l sinθ ]- [ (m/2) g (l/2) sinθ ] .................................. (1)

where F_b is bouyant force from liquid , F_g is force due to gravity ( weight ) .

Force in upward direction is consdered as positive . Downward force is considered as negative .

( F_b - F_g ) = { ρ_l A [ (h/2) + y ] g } - [ ρ_c A h g ] .......................(2)

where ρ is density , subscript l denotes liquid and subscript c denotes cylinder. A is cross section area of cylinder and g is accelkeration due to gravity.

It is given that ρ_l= 2 ρ_c .............................(3)

Using eqn.(3) , we simplify eqn.(2) as

( F_b - F_g ) = 2 ρ_c y A g = 2 ( ρ_c h A ) (y/h) g = 2 m (y/h) g

Total torque from eqn.(1) is written as

Total torque τ = { [ 2 m (y/h) g l ] - [ (m/2) g (l/2) ]} sinθ = [ 2 (y/h) - (1/4) ] mgl sinθ

By Newton's law , Total torque τ = I α , where I = (1/3) m l² is moment of inertia of rod about its end and α is angular acceleration .

τ = [ 2 (y/h) - (1/4) ] mgl sinθ = (1/3) m l² ( d²θ / dt² )

Above expression is written as

( d²θ / dt² ) = - { ( 3/ l ) [ (1/4) - 2 (y/h) ] g } sinθ

Angular frequeny for small amlitude of oscillation can be written as

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Answered by Thiyagarajan K | 10 Apr, 2022, 14:55: PM

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