M, a solid cylinder (M=2.39 kg, R=0.121 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cy

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M A Solid Cylinder M 2 39 Kg R 0 121 M Pivots On A Thin Fixed Frictionless Bearing A String Wrapped Around The Cy 1 (83.78 KiB) Viewed 47 times

M, a solid cylinder (M=2.39 kg, R=0.121 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 0.830 kg mass, i.e., F = 8.142 N. Calculate the angular acceleration of the cylinder. R M F 5.63x10¹ rad/s^2 You are correct. Your receipt no. is 164-8014 If instead of the force an actual mass m = 0.830 kg is hung from the string, find the angular acceleration of the cylinder. m Previous Tries 56.253rad/s 2 Hint: The tension in the string induces the torque in both this part and the first part. The tension is not equal to mg! If it were, the mass would not accelerate downward. Determine all of the forces acting on the mass, then apply Newton's second law and solve for the tension, and apply it to Newton's second law of rotational motion. Submit Answer Incorrect. Tries 5/12 Previous Tries How far does m travel downward between 0.570 s and 0.770 s after the motion begins? Submit Answer Tries 0/12 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 0.494 m in a time of 0.510 s. Find Iem of the new cylinder. Submit Answer Tries 0/12