M, a solid cylinder (M=1.95 kg, R=0.113 m) pivots on a thin, fixed, frictionless

Question

M, a solid cylinder (M=1.95 kg, R=0.113 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 0.750 kg mass, i.e., F = 7.357 N. Calculate the angular acceleration of the cylinder.

C)How far does m travel downward between 0.710 s and 0.910 s after the motion begins?   

D)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 0.367 m in a time of 0.470 s. Find Icm of the new cylinder.

Explanation / Answer

momenty of inertia of solid cylinder is I = 0.5*m*r^2 = 0.5*1.95*0.113*0.113 = 0.012449775 kg-m^2

relation between torque and angular accelaration

T = I*alpha

angular accelaration alpha = T/I = (r*F)/I = (0.113*7.3575)/0.012449775 = 66.78 rad/s^2

B) alpha= 66.78 rad/s^2

C) linear accelaration a = r*alpha = 0.113*66.78 = 7.54 m/s^2

the mass m falls with the constnt accelaration

time take is 0.91-0.71 = 0.2 s

initial speed is u = 0 m/s

then apply s = (u*t) +(0.5*a*t^2)
S = 0.5*7.54*2*2 = 15.08 mis the answer for C)

D) new accelaration is a = (2*S)/t^2 = (2*0.367)/(0.47*0.47) = 3.32 m/s^2

angular accelaration is a/r = 3.32/0.113 =29.4 rad/s^2

Torque is T = (0.113*7.3575) = 0.8313975 N-m

Inew = T/alpha = 0.8313975/ 29.4 = 0.02827 kg-m^2

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