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Q.

A thin circular rod is supported in a vertical plane by a bracket at A. Attached to the bracket and loosely wound around the rod is a spring of constantk=40N/m and under formed length equal to arc of the circle AB. A 0.2kg collar ‘C’, not attached to the spring can slide without friction along the rod. The collar is released from rest at an angleθ with the vertical. Make the equation for minimum value of θ  for which the collar will pass through D and reach point A is βθ2=1+cosθ where β=

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a

answer is C.

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Detailed Solution

at point C

From law of conservation of energy

T.EC=P.E+K.E

=mgh+12kx2

=mgR(1cosθ)+12k(Rθ)2

=(0.2)(9.81)(0.3)(1cosθ)+12(40)(0.3θ)2

=1.8θ2+0.5886(1cosθ)J

T.EP=(K.E+P.E)D

=0+2mgR

=2(1.962)(0.3)

=1.772J

From law of conservation of energy

T.EC=T.EP

1.8θ2+0.5886(1cosθ)=1.1772

θ=07552(rad)

=0.7552(rad)

θ=43.1°

In question

βθ2=1+cosθ

θ(0.7552)2=1+cos(43.1°)

β(0.57032)=1+0.7301

β=1.73010.57032=3.0335

β=3

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