Mechanics> angular-velocity...

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??

Shuvam Shukla , 11 Years ago

Grade 12

3 Answers

SOURAV MISHRA

Last Activity: 11 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

Last Activity: 6 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

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

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