Solve Maximum Speed Equation: I (of P) = I (of o) + Md*d

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The equation I (of P) = I (of o) + Md*d is central to the discussion, with users struggling to apply it to find answers related to potential energy changes when the center of mass is at the bottom. There is confusion about the conservation of energy principle and its application in the context of the problem. A user attempts to calculate kinetic energy before and after a change in rotation, leading to the conclusion that ω2 = ω1√3. Despite these calculations, the user expresses frustration over not obtaining a specific value for the answer. The conversation highlights the challenges in understanding and applying rotational dynamics and energy conservation principles.
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When the center of mass is at the bottom what is its change in potential energy?

Where did the PE go?
 
LowlyPion said:
When the center of mass is at the bottom what is its change in potential energy?

Where did the PE go?

so do i use energy conservation...

please tell me soon its due in an half our :(
 
dsptl said:
so do i use energy conservation...

please tell me soon its due in an half our :(

Why not use conservation of energy?
 
LowlyPion said:
Why not use conservation of energy?

huh?
 
my 2nd trial:

I displaced= I shape axis + mR2

For the cylinder Ic is
Ic = 0.5 m R2

Ke= 0.5 I ω^2
the energy must be concerved when the center of rotation is changed from P to O.
We have
Ke(berore) =0.5 [0.5 m R2 + mR2] ω1^2 = 0.25[ 1+2] mR2ω1^2 = (3/4)mR2ω1^2
Ke(after) =0.5 [0.5 m R2 ] ω2^2 = (1/4) mR2ω2^2 we have

(3/4)mR2ω1^2 = (1/4) mR2ω2^2
ω2= √[(3/4)mR2ω1^2 /(1/4) mR2 ]=ω1√ 3

ω2= ω1√ 3


now tell me whts wrong here...i am not getting any value for ans??
 
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