M, a solid cylinder (M=1.55 kg, R=0.127 m) pivots on a thin, fixed, frictionless

Question

M, a solid cylinder (M=1.55 kg, R=0.127 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 0.850 kg mass, i.e., F = 8.338 N. Calculate the angular acceleration of the cylinder.

already calculated: 8.47E1 rad/s^2

1.) If instead of the force F an actual mass m = 0.850 kg is hung from the string, find the angular acceleration of the cylinder.

3.) 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 0.496 m in a time of 0.470 s. Find Icm of the new cylinder.

I already calculated the angular acceleration of the cylinder, but i am having trouble with the remainder of this problem, please help!

Explanation / Answer

Ans:- M= 1.55kg, R= 0.127m,m= 0.850kg ,F = 8.338N

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