1. The problem statement, all variables and given/known data A uniform rod is pivoted at its center and a small weight of mass M = 5.07 kg is rigidly attached to one end. You may neglect the size of the weight and assume it is located right at the end of the rod. a) What vertical force, Fv, must be applied to the opposite end to keep the rod at an angle of q = 37° to the horizontal? Answer to part a) I calculated the answer to part a and came up with 49.74 N b) The vertical force is now replaced by a horizontal force, Fh. How large must this be to hold the rod at the q = 37° angle? Answer: 66.0 N c) Now suppose that the rod has length L = 6.7 m and mass mrod = 10.9 kg. Suppose also that there are no external forces applied (i.e. Fh = Fv = 0). The system is released from rest at the q = 37° angle. What is the angular acceleration just after it is released (in rad/s^2)? This part I do not know how to answer. I attemped it using the equation sum of torques= (moment of inertia I)* (angular acceleration) 2. Relevant equations 3. The attempt at a solution Sum of torques= I* (angular acceleration) The moment of inertia of a rod is (1/12) ML^2, where the mass of the rod is 10.9 kg. And L is 6.7 m. [ (1/12)(10.9kg)(6.7m)^2 = 40.775 kh*m^2.] I added this to the moment of inertia of the weight which is mr^2 [5.07m* (6.7/2)^2 = 56.9] I combined the two to get the moment of inertia(total) = 97.673 then for torque= rF, I used F= mg= (10.9+5.070(9.81)), which came out as 156.67. I multiplied that number by r= 3.35, and came out with 524.83. so to find angular acceleration: 524.83/97.67= 5.37. However, this turned out wrong. My way of calculating the moment of inertia might be wrong? Somebody please help me. Part d) What is the angular velocity when the rod is vertical? It says to use conservation of energy but I'm not sure where to start. Thank you and help will be much appreciated.