How far does m travel downward between 0.610 s and 0.810 s after the motion begi

Question

How far does m travel downward between 0.610 s and 0.810 s after the motion begins?

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 0.607 m in a time of 0.570 s. Find Icm of the new cylinder.

M, a solid cylinder (M=1.51 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 0.650 kg mass, i.e., F = 6.377 N. Calculate the angular acceleration of the cylinder. How far does m travel downward between 0.610 s and 0.810 s after the motion begins? 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 0.607 m in a time of 0.570 s. Find Icm of the new cylinder. If instead of the force F an actual mass m = 0.650 kg is hung from the string, find the angular acceleration of the cylinder.

Explanation / Answer

torque = r x T   = 0.137 x 0.650 x 9.81 = 0.874
torque = I *alpha
0.874 = (1.51 x 0.137^2 /2 ) * alpha
alpha = 61.68 rad/s2

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on cyliner :
rT    = I *alpha  

=> rT = (Mr^2 /2) (a/r)

=> T = Ma/2
on mass :
mg - T = ma     
add them
mg - Ma/2 = ma

mg = a( m + M/2 )

a = 4.54 m/s2
alpha = a/r = 4.54 / 0.137 = 33.13 m/s2

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a = alpha.r =33.13 x 0.137 = 4.54 m/s2

d = a(t2^2 - t1^2 ) /2    = 4.54 x (0.810^2 - 0.610^2 ) / 2 = 0.64 m

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0.607 = a (0.570^2) /2
a = 3.74 m/s2
alpha = a/r = 27.27 m/s2
mgr = (mr^2 + I ) alpha
0.650 x 9.81 x 0.137 = ( 0.650 x 0.137^2 + I) 27.27
I = 0.02 kg.m^2

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