38. A uniform rod of length 'l' is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed ω the rod makes an angle θ with it (see figure). To find θ equate the rate of change of angular momentum (direction going into the paper) (𝑚𝑙^2)/12 𝜔^2sin θ cos θ about the centre of mass (CM) to the torque provided by the horizontal and vertical forces FHFH and FV about the CM. The value of θ is such that: (a) cosθ=2g/3l𝜔^2 (b) cosθ=g/2l𝜔^2 (c) cosθ=g/l𝜔^2 (d) cosθ=3g/2l𝜔^2
option : d
38. A uniform rod of length 'l' is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed ω the rod makes an angle θ with it (see figure). To find θ equate the rate of change of angular momentum (direction going into the paper) (𝑚𝑙^2)/12 𝜔^2sin θ cos θ about the centre of mass (CM) to the torque provided by the horizontal and vertical forces FHFH and FV about the CM. The value of θ is such that: (a) cosθ=2g/3l𝜔^2 (b) cosθ=g/2l𝜔^2 (c) cosθ=g/l𝜔^2 (d) cosθ=3g/2l𝜔^2
38. A uniform rod of length 'l' is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed ω the rod makes an angle θ with it (see figure). To find θ equate the rate of change of angular momentum (direction going into the paper) (𝑚𝑙^2)/12 𝜔^2sin θ cos θ about the centre of mass (CM) to the torque provided by the horizontal and vertical forces FHFH and FV about the CM. The value of θ is such that: (a) cosθ=2g/3l𝜔^2 (b) cosθ=g/2l𝜔^2 (c) cosθ=g/l𝜔^2 (d) cosθ=3g/2l𝜔^2
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