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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

Rupali Sehgal
21 Points
7 years ago
when the disc is inclined at angle theta,
Mg (wt) acts downwards at R/2 distance from P ie at C. Normal acts at the point of pivot. 
Net torque about P= I (alpha)
MgRcos(theta) + N(0) = (1/2) MR^2 (alpha)
alpha = 2gCos (theta) / R

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