M, a solid cylinder (M = 1.83 kg, R=0.137 m) pivots on a thin, fixed, frictionle

Question

M, a solid cylinder (M = 1.83 kg, R=0.137 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. 6.50 times 10^1 rad/s^2 If instead of the force F an actual mass m = 0.830 kg is hung from the string, find the angular acceleration of the cylinder. If instead of the force F an actual mass m = 0.830 kg is hung from the string, find the angular acceleration of the cylinder.

Explanation / Answer

part 1:

moment of inertia of solid cylinder=0.5*mass*radius^2=17.1736*10^(-3) kg.m^2

value of g=8.142/0.83=9.81 m/s^2

F=8.142 N

then angular acceleration=torque/moment of inertia

=F*radius/moment of inertia=64.9517 rad/s^2

part 2:

let tension in the string is T.

let linear acceleration is a m/s^2.

force equation for mass m:

m*g-T=m*a...(1)

force equation for mass M:

T*R=17.1736*10^(-3)*(a/R)

==>T=0.915*a...(2)

using equation 2 in equation 1:

0.83*9.81-0.915*a=0.83*a

==>a=4.666 m/s^2

then angular acceleration=a/R=34.06 rad/s^2

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