- #1
Pepachin
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Hello!, I am trying to solve this problem of the book beer vector mechanics for engineers statics and dynamics, I`ll be very glad if you can help me with this
http://i.imgur.com/pFkUfuv.png (Image of the problem)
Each of the two identical slender bars shown has a length L. Knowing that the system is released from rest when the bars are horizontal, calculate the equations for angular velocity(ω) of rod AB and velocity of point D for values of θ from 0 to 90 degrees.
Here are some formulas that might help:
I= 1/12 mL^2
Principle of conservation of energy T1 + V1 = T2+V2
V=mgh
T= (1/2)*Iω^2
V= ωxr
First I tried to solve as a mechanism of three bars but it didn´t work and I´m trying to solve it with moments of Inertia I= 1/12 mL^2 and kinematics but the problem doesn´t say the mass of the rods, I also tried to solve for the equations of the two positions (when the rods are horizontal and when the angle is 90) but I can not get to anywhere
Thanks for taking your time to read this.
http://i.imgur.com/pFkUfuv.png (Image of the problem)
Each of the two identical slender bars shown has a length L. Knowing that the system is released from rest when the bars are horizontal, calculate the equations for angular velocity(ω) of rod AB and velocity of point D for values of θ from 0 to 90 degrees.
Here are some formulas that might help:
I= 1/12 mL^2
Principle of conservation of energy T1 + V1 = T2+V2
V=mgh
T= (1/2)*Iω^2
V= ωxr
First I tried to solve as a mechanism of three bars but it didn´t work and I´m trying to solve it with moments of Inertia I= 1/12 mL^2 and kinematics but the problem doesn´t say the mass of the rods, I also tried to solve for the equations of the two positions (when the rods are horizontal and when the angle is 90) but I can not get to anywhere
Thanks for taking your time to read this.