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A uniform rod of length l is hinged at a distance of l/4 from left and is released from a horizontal position. The angular velocity of the bat as it passed the vertical position is??

A uniform rod of length l is hinged at a distance of l/4 from left and is released from a horizontal position. The angular velocity of the bat as it passed the vertical position is??

Grade:12

3 Answers

SOURAV MISHRA
37 Points
8 years ago

by parallel axes theorem the moment of inertia of the rod about the hinge is mL2/12 + m(L/4)2 = 7mL2/48.

when the rod reaches the vertical position the center of mass of the rod falls by L/4.

by conservation of energy we can write:

mgL/4 = (1/2)*I*w2       where I = moment of inertia = 7mL2/48 and w = angular velocity of the rod at the vertical position

this gives

w = (24g/7L)1/2.

 

Alok kumar virat
15 Points
3 years ago
In this situation we know that work energy theorem a energy nd applying this . This theorem say that work done by all forces should be change in kinetic energy. So 
Here work done by gravity = k(final)-k ( initial)
So mgh = kf - 0 (becoz just released  from rest)
So next here h should be centre of mass distance then  
Mg×l/4 = 1/2 I w^2 .....equation (1).( here w = omega
Now For find I 
I = inertia of rod and inertia of hing 
Ml^2/12 +mr^2 
 Ml^2/12 + m×(l/4)^2 = 7 ml^2/48 
Put value in equation 1 then 
Mg×l/4= 1/2×7ml^2/48×w^2
Solving this equation and find w 
And after solving w^2= 24g/7l  and w  = root under 24g/7l .
This is the final answer 
Thanku 
 Your Alok virat.. class 11
And again 
Yash Chourasiya
askIITians Faculty 256 Points
one year ago
Dear Student

By parallel axis Theorem, MOL of rod
about hinge is (I0) = 7mL2​/48

When rod reaches the vertical positionthe center of mass of rod falls byL/4
By consecration of energy
mg(L/4) = 1/2(I0​w2)
mgL/4 = 1/2×(7mL2​w2/48)
w = √24g/7L​​

I hope this answer will help you.
Thanks & Regards
Yash Chourasiya

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