A yo-yo has mass m and rotational inertia I. A light thin string is wrapped around the central part of the yo-yo that has radius. The string is attached to the ceiling as shown. The yo-yo is released from rest and begins to descend as the string unwraps. The positive "y" direction is downward, as indicated by the arrow on the diagram. Part 1. If the force of tension in the string is T, what is the correct expression that results from applying Newton's 2nd law on the motion of the center of mass of the yo-yo? Remember that downwards is treated as the positive direction in this problem. Correct: mg-T=ma Incorrect T+mg=ma Incorrect Zero Incorrect mg-Tr=ma Incorrect mg+Tr=ma Incorrect Tr-mg=ma You are correct. Your receipt no. is 149-8234 Previous Tries Part 2. With respect to the axis passing through the center of the yo-yo, what is the magnitude of the net torque acting on the yo-yo? Take clockwise rotation to be the positive direction. (mg-T)r (mg+T)r m, I 8 750 o mgr 2Tr O2mgr Submit Answer Incorrect. Tries 2/3 Previous Tries Part 3. What is the relationship between the translational acceleration of the yo-yo a and its angular acceleration ? O a=ar a = a/r O a=2ar o a=2x/r 0 a=ar/2 a=/(2r) Submit Answer Tries 0/3 Part 4. What is the downward acceleration of the yo-yo? a=g(mr)/(mr+1) a=g(mr+1)/(mr) a=g(mr2)/(1) a=g(1)/(mr2) a=g(1)/(mr2+1) a=g(mr2+1)/1) Submit Answer Tries 0/3 Part 5. What is the magnitude of the force of tension in the string? Part 1 mg (mr2)/(mr2+1) T= mg(mr+1)/(mr) T=mg(mr)/(1) T= mg(1)/(mr) T= mg(1)/(mr2+I) T= mg(mr2+1)/I) Submit Answer Tries 0/3