Questions
A block of mass m = 1 kg slides with velocity on a frictionless horizontal surface and collides with a uniform vertical rod and sticks to it as shown. The rod is pivoted about O and swings as a result of the collision making angle before momentarily coming to rest. If the rod has mass M = 2 kg, and length l = 1 m, the value of is approximately:
detailed solution
Correct option is C
Initial angular momentum of block about hinge before Collison = Final angular momentum of combine system after collision ⇒mvl=L ---(1)Now, using energy conservation Loss in rotational kinetic energy of combine system = Gain in potential energy of combine system⇒L22I=mgl1−cosθ+Mgl21−cosθ ∵cosθ=l−hl⇒h=l1−cosθ ⇒mvl22ml2+Ml23=m+M2gl1−cosθ ∵I=Iblock+Irod=ml2+Ml23& from 1⇒1×6×1221×12+2×123=1+2210×11−cosθ⇒cosθ=2350∵m=1kg, l=1m and M=2kg⇒θ=630Talk to our academic expert!
Similar Questions
The condition of equilibrium is satisfied when resultant of all forces acting on body vanishes and also no net torque acts on the body. However body can be different types of equilibrium as stable, unstable, neutral or may be having pure rotation when no resultant force acts on the body. Different conditions of equilibrium or rotational motion can be known by different types of energy it possesses. Column I mentions the state of body and column II mentions the type of energy it possesses. Match the correct options in two columns
COLUMN_I COLUMN_II
A) Body in stable equilibrium P) Rotational kinetic energy
B) Body in unstable equilibrium Q) Minimum potential
energy
C) Body in neutral equilibrium R)Maximum potential energy
D) Body in pure rotational motion S) Constant potential energy
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