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flower76
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I have figured out the first part regarding the equation, but I don't really know what to write for the second part. Intuitively I think the rod would swing down, because m1 is heavier, and it would have the greatest velocity when vertical. But I have no idea if I'm right, and if I am I can't give reasons why.
The question:
A rigid rod of mass M and length I, with masses m1 and m2 attached at the end of the rod, can rotate in a vertical plane about a frictionless pivot through its centre. At time t=0, the rod is held at an angle theta, as shown. Show that if the system is let go, the angular acceleration of the rod at t=0 is:
[tex]\alpha=\frac{2(m1-m2)gcos\theta}{L(M/3+m1+m2)}[/tex]
If m1>m2, for what value of [tex]\theta[/tex] is the angular velocity [tex]\omega[/tex] a maximum? Give reasons for your answer.
The diagram shown has a rod at about a 45 degree angle north east with m1 the mass on the bottom.
Help is much appreciated.
The question:
A rigid rod of mass M and length I, with masses m1 and m2 attached at the end of the rod, can rotate in a vertical plane about a frictionless pivot through its centre. At time t=0, the rod is held at an angle theta, as shown. Show that if the system is let go, the angular acceleration of the rod at t=0 is:
[tex]\alpha=\frac{2(m1-m2)gcos\theta}{L(M/3+m1+m2)}[/tex]
If m1>m2, for what value of [tex]\theta[/tex] is the angular velocity [tex]\omega[/tex] a maximum? Give reasons for your answer.
The diagram shown has a rod at about a 45 degree angle north east with m1 the mass on the bottom.
Help is much appreciated.