How far does m travel downward between 0.510 s and 0.710 s after the motion begi

Question

How far does m travel downward between 0.510 s and 0.710 s after the motion begins?M, a solid cylinder (M=1.99 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 0.670 kg mass, i.e., F = 6.573 N. Calculate the angular acceleration of the cylinder.

If instead of the force F an actual mass m = 0.670 kg is hung from the string, find the angular acceleration of the cylinder.

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 0.378 m in a time of 0.510 s. Find Icm of the new cylinder.

Explanation / Answer

part a )

t = rF*sin theta

t = I*alpha

rF sin theta = I*alpha

sintheta = 1

alpha = r*F/I

I for solid cylinder = 1/2 Mr^2

alpha = 2F/Mr

alpha = 2 * 6.573 / 1.99 * 0.127

alpha = 52.02 rad/s^2

a = 52.02 * 0.127 = 6.61 m/s^2

s = 0 + 1/2 * 6.61 * (0.510)^2 = 0.8596 m

s = 0 + 1/2 * 6.61 * ( 0.710)^2 = 1.6661 m

ds = 0.8014 m

part b )

T = mg - ma

a = r * alpha

T = mg - m*alpha*r

t = I*alpha

R*T = 1/2*MR^2 *alpha

alpha = 2T/MR

alpha = 2mg / R(M+2m)

alpha = 31.05 rad/s^2

part c )

s = ut + 1/2 at^2

0.378 = 1/2 * a *(0.510)^2

a =2.91 m/s^2

alpha = a/r = 2.91/0.127 = 22.91 rad/s^2

t = r*F = 0.127 * 0.670 * 9.8 = 0.8339 N-m

I = t/alpha

I = 0.8339/22.91 = 0.0364 kg .m^2

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