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Grade: 11

                        

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?
 

4 years ago

Answers : (1)

Vikas TU
14136 Points
							
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)
4 years ago
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