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=40 N/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=

Question

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=40 N/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=

Infinity Learn · Accepted Answer

at point C

From law of conservation of energy

$T {.E}_{C} = P .E + K .E$

$= mgh + \frac{1}{2} k x^{2}$

$= mgR (1 - \cos θ) + \frac{1}{2} k {(R θ)}^{2}$

$= (0.2) (9.81) (0.3) (1 - \cos θ) + \frac{1}{2} (40) {(0.3 θ)}^{2}$

$= 1.8 θ^{2} + 0.5886 (1 - \cos θ) J$

$T {.E}_{P} = {(K .E + P .E)}_{D}$

$= 0 + 2 mgR$

$= 2 (1.962) (0.3)$

$= 1.772 J$

From law of conservation of energy

$T {.E}_{C} = T {.E}_{P}$

$1.8 θ^{2} + 0.5886 (1 - \cos θ) = 1.1772$

$θ = 0 - 7552 (rad)$

$= 0.7552 (rad)$

$θ = 43.1 °$

In question

$β θ^{2} = 1 + \cos θ$

$θ {(0.7552)}^{2} = 1 + \cos (43.1 °)$

$β (0.57032) = 1 + 0.7301$

$β = \frac{1.7301}{0.57032} = 3.0335$

$β = 3$

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