IIT JEE , AIEEE Forum - Physics , Mechanics

kishan12

A rod of mass 1 Kg(uniformly distributed) length 1 m is pivoted at its centre and 2 masses of 5kg and 2kg are hung from the ends as shown in figure.Find the initial angular acceleration of the rod assuming that it was horizontal in the begining.Also find the tension in the supports to the blocks of mass 2Kg and 5kg.

joyfrancis

Force due to 5kg block = 5g downwards

.: torque of this force about center = 5g * (1/2) (clockwise)

&torque due to 2kg block = g (anticlockwise)

Net torque at the center = 5g/2-g=3g/2

Moment of inertia of the system about center = moi of 5kg + moi of 2kg

= 5*(1/2)² + 2(1/2)

= 5/4+2/4 = 7/4

Net torque = I

3g/2 = 7/4 * (

)

.:

= 8.4 rad/sec

kishan12

Hey joy look at quest...
its actually Q:32 and not 31 which u solved..

rooney

In the question, mass of rod is given to be 1 Kg.

So, I = 7/4 + 1x(1)^2 / 12
So, you get new value of Alpha.

As for tensions, let us consider the 5kg mass.
Its acceleration is = 0.5 m x Alpha = A

So, (weight of 5kg mass) - Tension = 5 x A
You get Tension from this equation. Same with the other mass.

Regards
Rooney

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