Question

A rod of length L=103 m and density ρ is pivoted about its one end is submerged in liquid of density 2ρ and held horizontal as shown. Now if the rod is released from rest, the angular speed in rad/s when it becomes vertical is [Answer upto two decimal points][Take g=10 m/s2]

Solution

As the density of rod is less than the liquid, the buoyancy force B will be acting upwards on the rod and the weight of the rod acts downwards. Let M, V be the mass and volume of the rod respectively. Hence using the work-energy theorem.  ⇒BL2−MgL2=12Ioω2 where Io is the moment of Inertia about O, Io=13ML2⇒(2ρVg)L2−(ρVg)L2=12M3L2ω2⇒VρgL2=12(ρV)3L2ω2⇒ω=√3gL=√3×10×310=3 rad/s  Co-Curriculars

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