A uniform drawbridge 8.00 m long is attached to the

Solution 86P Introduction The linear acceleration in this case is the acceleration due to gravity. We have to find out the tangential acceleration at the given point and then we can calculate the angular acceleration. Step 1 (a) We can consider that the gravity is acting from the center of mass. Since the center of mass will be at the center of the drawbridge, the distance from the hinge to the center of mass is (8.00 m)/2 = 4.00 m. The gravity always acts downwards, if the angle of the drawbridge is , then the tangential acceleration of the center of mass is given by a t gsin Now the angular acceleration is given by = a t = grsin Now from the given problem we know that r = 4.00 m, = 60°. We also know that the acceleration due to gravity is g = 9.8 m/s . Hence we angular acceleration is given by = (9.8 m/s )(4.00 m)sin60° 33.9 rad/s 2