Angular acceleration of a rod - 2

[Vibhor]

Homework Statement

Homework Equations

The Attempt at a Solution

[/B]
Initially the rod is in rotational equilibrium , so net torque about CM is zero .

From torque equation about CM , we get

Tension T in the left string = Force F (kx) in the spring

Doing a force balance gives us T+F=Mg

2F=Mg

F=Mg/2

Now when the string is cut , force equation in vertical direction is Mg - F = Ma

Torque equation about the CM is F(L/2) = (ML²/12)α

F = Mg/2 since force exerted by spring remains same just after right string is cut .

Using value of F in the above two eqs gives a = g/2 and α = (3g/L)

Have I approached it correctly ?

 

[kuruman]

Have I approached it correctly ?

No. When the cord is cut, the rod will rotate about its end at A. It is best to do your calculation about that end. Then you will see that the force exerted by the spring changes when the cord is cut.

 

[Vibhor]

No. When the cord is cut, the rod will rotate about its end at A. It is best to do your calculation about that end. Then you will see that the force exerted by the spring changes when the cord is cut.

Just after the cord is cut , the spring length doesn't change instantaneously , which means force exerted by spring doesn't change immediately when the cord is cut .

I don't see how spring force is different before and just after cord is cut .

 

[kuruman]

Just after the cord is cut , the spring length doesn't change instantaneously , which means force exerted by spring doesn't change immediately when the cord is cut .

That is true. I was thinking about another situation when a stated that the force changes. For this problem, the angular acceleration for rotations about the CM is the one that makes sense. Your solution looks good.

 

[Vibhor]

Thanks