Sliding Mass on Fixed Rod: Analyzing Angular Momentum

[neelakash]

Homework Statement

A smooth ring of mass m can slide on a fixed horizontal rod.A string tied to the ring passes over a fixed pulley and carries a mass M(<m).At an instant the angle between the rod and the string is θ.Show that if the ring slides with a speed v,the block descends with a speed v cosθ.With what acceleration will the ring start to move if the system is released from rest at θ=30*?

I have attached the figure in a pdf file so that you may see it

Homework Equations

The Attempt at a Solution

I take z axis downwards,x axis rightwards.

The force equations:T cosθ=m D²x
Mg-T=M D²z

Now,we are to find the constraint equation.

I got this:(using the length conservation)

√[x²+c²]+z=L...c is a const
Differentiating twice w.r.t. t we get: Dz=-Dx (cosθ)


What is annoying me is the (-)ve sign before the ansswer.

Somehow this is not correct.Because,in the next part we require this result:

D²z=-D²x cosθ+0(initially Dx=0)

This gives an error in the final answer.

Can anyone please help?

 

[Doc Al]

The force equations:T cosθ=m D²x
Mg-T=M D²z

In your constraint equation, it looks like you use "x" to represent the horizontal distance between sliding mass and pulley. x decreases as the mass slides. So the acceleration of the sliding mass should be -D²x.

 

[neelakash]

Quote:
Originally Posted by neelakash
The force equations:T cosθ=m D²x
Mg-T=M D²z

In your constraint equation, it looks like you use "x" to represent the horizontal distance between sliding mass and pulley. x decreases as the mass slides. So the acceleration of the sliding mass should be -D²x.

Exactly! was not careful to write the force equations...it should be consistent with the constraint equation's sign convention.
I got the correct answer!

Thank you very much.