M, a solid cylinder (M=1.47 kg, R=0.117 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cy

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M A Solid Cylinder M 1 47 Kg R 0 117 M Pivots On A Thin Fixed Frictionless Bearing A String Wrapped Around The Cy 1 (13.91 KiB) Viewed 34 times

M A Solid Cylinder M 1 47 Kg R 0 117 M Pivots On A Thin Fixed Frictionless Bearing A String Wrapped Around The Cy 2 (27.91 KiB) Viewed 34 times

M, a solid cylinder (M=1.47 kg, 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. M F 9.92x10¹ rad/s^2 You are correct. Your receipt no. is 164-1422 Previous Tries
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. M m 4.54x10¹ rad/s^2 You are correct. Your receipt no. is 164-8189 Previous Tries How far does m travel downward between 0.390 s and 0.590 s after the motion begins? 5.21x10-¹ m You are correct. Your receipt no. is 164-60 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 0.438 m in a time of 0.450 s. Find Im of the new cylinder. 0.1280kg 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,