QUESTION: The 1.25kg slender rod AB is attached to spring BC which has an unstre

Question

QUESTION: The 1.25kg slender rod AB is attached to spring BC which has an unstretched length of 1.2m. If the rod is released from rest when Ø=30°, determine the angular velocity of the rod the instant the spring becomes unstretched.

So I'll put down what I've calculated so far. I know I have done something wrong as the answer isn't coming together when I put it all into the conservation of energy equation T1 + V1 = T2 + V2.

If someone could point out where I've went wrong that would be a great :)

BC = 2.318m (when Ø=30°)

VelocityG (vG)

vG = 0.6w

IG= 1/12ml^2 + 1/2ml                 ( I know somethings wrong here)

= 1.5 + 7.5

= 9

T2 = 1/2mv^2 + 1/2IGw^2

     = 1/2*12.5*(.6w)^2 + 1/2*9*w^2

     = 6.75w^2

Vg = mgy

Vg1 = 12.5*9.81*0       (G as the datum)

      = 0

Vg2 = 12.5*9.81*(1.2^2 - 0.6^2)^0.5

       = 90.6444

Ve = 0.5ks^2

Ve1 = 0.5*80*(2.318-1.2)^2

      = 49.996

Ve2 = 0

T1 = 0

T1 + V1 = T2 + V2

0 + 49.996 = 90.6444 +6.75w^2

??? ^^^^ that's definitely not right.

THE ANSWER SHOULD BE w = 2.776rad/s

Explanation / Answer

IG= 1/12ml^2 + 1/2ml this is wrong. first of all where is the point G? and secondly the above equation is dimensionally incorrect. I of rod about any end is 1/3ml^2 I of rod about central axis is 1/12ml^2 if the axis is at r distance from these axis.. use I = I about axis + mr^2

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