A uniform drawbridge 8.00 m long is attached to the roadway by a frictionless hinge at one end, and it can be raised by a cable attached to the other end. The bridge is at rest, suspended at 60.0° above the horizontal, when the cable suddenly breaks. (a) Find the angular acceleration of the drawbridge just after the cable breaks. (Gravity behaves as though it all acts at the center of mass.) (b) Could you use the equation ??? = ???0 + ??t? to calculate the angular speed of the drawbridge at a later time? Explain why. (c) What is the angular speed of the drawbridge as it becomes horizontal?

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 Step 2 (b) The given equation of the final angular velocity = +0t Is valid only if the angular acceleration is constant. But as we have seen for the drawbridge, the angular acceleration is not constant, instead it varies with sine of the angle made by the drawbridge with the vertical. Hence the given equation will not be true.