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??

Question

SOURAV MISHRA · Answer

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 · Answer

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 · Answer

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(I₀w²)
mgL/4 = 1/2×(7mL²w²/48)
w = √24g/7L

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

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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??

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