A uniform rod is rotated about it\'s an axis through one of its ends. The rod go

Question

A uniform rod is rotated about it's an axis through one of its ends. The rod goes from rest to an angular speed of 19 rad/s in a time of 0.55s with a constant angular acceleration. The rod has length L=120 cm and a mass of 1.35 kg.

A.) Draw a sketch showing the rod rotating about the axis.

B.) Calculate the angular acceleration during the 0.55s interval

C.) Calculate the net torgue on the rod during the time 0.55s interval

E.) Calculate the speed of a point halfway between the end of the rod and its rotation axis at a time od 0.20s after the applied torque is removed?

Explanation / Answer

B)
w = w0 + alpha*t
19 = 0 + alpha*0.55
alpha = 34.545 rad/s^2

C)
MOI = ML^2 /3
= 1.35*1.2^2 /3
= 0.648 kg-m^2

Torque = MOI*alpha
= 34.545*0.648
= 22.385 Nm

d)
During accleration phase angle turned is given by
theta = w0*t + 1/2*alpha*t^2
theta = 0 + 1/2*34.545*0.55^2
theta = 5.224 radians

During deceleration phase, angle trned is given by
0 = w^2 - 2*alpha2*theta2
0 = 19^2 - 2*8.92*theta2
theta2 = 20.235 radians

Total rotation = 5.224 + 20.235 = 25.459 radians (= 1458.7 degrees)

e)
Using w2 = w1 - alpha2*t we get
w2 = 19 - 8.92*0.2
w2 = 17.216 rad/s

v2 = r2*w2
= (L/2)*w2
= (1.2 / 2)*17.216
= 10.3296 m/s

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