- #1
dyn
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- Homework Statement
- A thin uniform rod of length 2a and mass m stands vertically on a frictionless horizontal table. The rod is slightly disturbed and falls down. What is its angular speed when it hits the table ?
- Relevant Equations
- Conservation of mechanical energy , PE = mgh , KE = (1/2)mv^2 + (1/2) Iw^2
I have seen the solution and understand it. The solution defines θ to be the angle between the falling rod and the table. It then equates initial PE to the final (PE+KE) where the final PE=0 and the final KE to be (1/2) Iω2+ (1/2) ma2ω2 to finally obtain ω = √(3g/2a)
But i would like to know why method doesn't work. I took the Centre of Mass (CM) to only be acted upon by gravity so it descends in a straight line with constant velocity g , using "suvat" equations , first gives the time of descent as t = √(2a/g) which means the linear velocity of the CM when it hits is √(2ag). This gives a final linear KE which cancels with the initial PE and so gives ω=0 which is obviously wrong. What is wrong with my reasoning ?
Thanks
But i would like to know why method doesn't work. I took the Centre of Mass (CM) to only be acted upon by gravity so it descends in a straight line with constant velocity g , using "suvat" equations , first gives the time of descent as t = √(2a/g) which means the linear velocity of the CM when it hits is √(2ag). This gives a final linear KE which cancels with the initial PE and so gives ω=0 which is obviously wrong. What is wrong with my reasoning ?
Thanks