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a uniform disc of mass m and radius r is pivoted at a point P on its rim and is free to rotate in the vertical plane.the centre C of disc is initially in horizontal position with respect to P.if it is released fromn this position ,what will be its angular acceleration when the line PC is inclined to the horizontal at an angle theta?

 
a uniform disc of mass  m and radius r is pivoted at a point P on its rim and is free to rotate in the vertical plane.the centre C  of disc is initially in horizontal position with respect to P.if it is released fromn this position ,what will be its angular acceleration when the line PC is inclined to the horizontal at an angle theta?
 

Grade:11

1 Answers

Vikas TU
14149 Points
7 years ago
Moment of inertia about the point P is:
I = Icom + md^2
And torque about P point we get,
(mgcosthetha)*d = Ia
a = (mgcosthetha)*d/I = > (mgcosthetha)*d/(0.5mr^2 + md^2)
a = 2gdcosthetha/(r^2 + 2d^2)

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