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Rotational Kinetic energy and Rotational Inertia

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The total rotational inertia of the three rods is :
I = I
1
+ I
2
+ I
3
= 0 +
𝒎𝑳
𝟐
𝟑
+ mL
2
=
𝟒
𝟑
mL
2
The center of mass is located in the middle of rod2 , hence
x
cm
= L/2
So when the plane of the H is vertical , the center of mass will be falling a distance of L/2.

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The total rotational inertia of the three rods is : I = I1 + I2 + I3 =0+ 𝒎𝑳𝟐 𝟑 + mL2 = 𝟒 𝟑 mL2 The center of mass is located in the middle of rod2 , hence xcm = L/2 So when the plane of the H is vertical , the center of mass will be falling a distance of L/2. From conservation of energy : PE (top) = KE (bottom) 3mg(L / 2) = (1/2) I ω2 Hence I ω2 = 3mgL So ( 𝟒 𝟑 mL2 ) ω2 = 3mgL Hence ω2 = 𝟗𝒈 𝟒𝑳 So ω=√ 𝟗𝒈 𝟒𝑳 rad/s Name: Rotational Kinetic energy and Rotational Inertia Description: This document contains a practice questio ...
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