angular acceleration, net torque, applied torque, work?

Question

A uniform rod of mass 1.1 kg and length 46 cm is constrained to rotate on an axis about its center. An unknown uniform torque is applied to the rod as it rotates through .28 radians from rest, which requires .8 seconds. The applied torque is then removed and, coasting only under the influence of friction, the rod comes to rest after rotating through 1.6 radians, which requires 5 seconds.
• Determine the angular acceleration of the moving rod before and after the torque is applied.
• Determine the net torque for each phase of the motion.
• What is the applied torque?
• How much work is done by the net torque, by the applied torque and by friction during the first phase of motion?

Answer 1

T=I*a

and for phase I
th=.5*a*t^2 (in this case)

.28=.5*a*.8^2
a=2*.28/.8^2
a=0.875 rad/s^2

I=1/3*m*L^2
I=1/3*1.1*.46^2
I=0.0776

Phase I net torque is
T=0.068 Nm

For phase 2

1.6=.5*a*25

a=2*1.6/25
a(friction)=0.128 Nm
T(friction)=0.0776*.128
T(friction)=0.00993 Nm

The applied torque in phase I is
0.0776*(a(applied)-0.128)=0.0776*0.875
(a(applied)=0.128+0.875
=1.003
T(applied) is 0.0778 Nm

The work done by applied
0.778*.28
0.218 J

and by friction
0.00993*.28
0.00278 J

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