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(a) The shaft is at rest

(b) The shaft is rotating in a horizontal plane about its center at .04 rev/s

(c) The shaft is rotating in a horizontal plane about its center at .20 rev/s

(d) At what rate must the shaft rotate in order that it may be supported at one end only?

My attempt (a) w = 5 rev/s = 10*PI rad/s. moment of inertia I=m*r^2 = 5*(.5)^2. Let the force in each hand =p; torque = r*p therefore the torque (T) = 4.9N.m

(b) If the shaft rotates at .04 rev/s, then the wheel rotates at 5.04 rev/s

(d)the requirement is that the resultant vertical force = zero i.e., p - m*g = 0 therefore p (the normal reaction force) =5kg*9.8m/s/s. T=I*(alpha) = (alpha) = T/I=9.8/(5*.25)=7.84; also

**t**=d

**L**/dt. Any help in doing parts a,b,c and d would be welcome, many thanks.