Question

M, a solid cylinder (M=1.55 kq, R=0.117 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force F which equals the weight of a 0.870 kg mass, i.e., F = 8.535 N. Calculate the angular acceleration of the cylinder. 9.41x101 rad/s^2 You are correct. 65.4779 Your receipt no. is 155-4279 Previous Tries vous If instead of the force F an actual mass m = 0.870 kg is hung from the string, find the angular acceleration of the cylinder. lo lol 4.43x101 rad/s^2 You are correct. Your receipt no. is 155-8793 Previous Tries How far does m travel downward between 0.390 s and 0.590 s after the motion begins? 5.08x10-1 m You are correct. Your receipt no. is 155-3774 Previous Tries The cylinder is changed to one with the same mass and radius, but a different moment of inertia. Starting from rest, the mass now moves a distance 0.634 m in a time of 0.570 S. Find Icm of the new cylinder. 2.99x10-2kg m 2 You know how to write an equation for the torque from the second part. You are given quantities to calculate the acceleration (review motion in one dimension if this is not obvious), and in this case you need to solve for the moment of inertia. Submit Answer Incorrect. Tries 2/12 Previous Tries

I need help with part D please! :)

Answer 1

to Torque ritmas n black It Inertia z cylinder ER = Radius . Tamg-ma z=1& .= rt Ia = long-ma) a= mgr . Dy to we have Pyett 63 a pps 270.634 par 134 ' we have at²=s Go 25 > 2x 0.034 € 10.57² = 3.90 0.87x9,8x0.117 0.87X0.117 + I 3.90 = 0112 I=0.018 Kg m2