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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
9 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
4 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
2 years 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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