Question

a rod of length l is hinged from one end . it is brought to a horizontal position and released. the angular velocity of the rod when it is vertical position is

Answer 1

Solving by help of Centre Of Mass

length of rod = l/2 (length at centre of mass)
so there body changes it state of motion
i.e potential energy = kinetic energy
Now, let mass of rod be M
MgL/2=1/2Iw²
MgL = (ML²/3)w² (by putting momentum of inertia of rod)
w=√3g/L
where w is angular velocity of rod
I is moment of inertia of rod

Answer 2

Answer:

First of all whole mass of rod will be assumed to be concentrated at its center for calculation of change in potential energy. Hence  

ΔPE= MgL /2

ΔPE=MgL/2

. Also, you need to consider kinetic energy of rigid body around hinge instead of considering it for point mass -  

KE= 1/ 2  I ω ^2

KE=1/2Iω^2

Hence,  

ΔPE= MgL /2

ΔKE=1/2IW^2

ΔPE=MgL/2=ΔKE=1/2Iω^2

I= M L^ 2/ 3

I=ML^2/3

ω

follows.

Explanation:

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