Question

Please just READ once.

Imagine a bucket filled with water. Now that bucket is tied to the ceiling with an elastic string with a torsional constant C. The bucket has some mass m. It is fillled with water upto a height, say h. We are considering it to be a perfect cylinder. Now a tap like structure is created at bottom of bucket such that water is ejected tangentially. Can you make a function for the initial time period of the damped SHM? Any values or constants please assume as a general case.

Answer 1

@JamesJ try this mann.

Answer 2

First, what is the equation of motion (the differential equation; don't solve it; just write it down) if the elastic string offered no resistance to the torque from the water flowing out?

Answer 3

The equation will be an equation in theta and theta'', where ' = d/dt.

Then we will need to layer on a term for the torsion. That I am not sure exactly how to do and would need to look some things up.

Answer 4

Once we have the complete second order ODE, we can write down a solution in terms of the coefficients; that last piece I can also do.

But I need you to figure out how torsion will impact theta, theta', ... I.e., what is the torque the the rotated string puts on the bucket?

Answer 5

Hi, James.
Totally missed the notification for this. Thanks for replying.
About the torque dueto the Water initially, That would be = 2apgh*(h)
Whre a is area of hole. p is density. Not so sure.
Annd. Now the trouble is in getting a relation between the angle rotated and the water emptied.
Because torque varies as h.
Thereore theta does.
Rstoring torque is well, T= 1/2 pi*(modulus of rigidity)* r^4 /l * theta.

Answer 6

How can we smplysubtract it?
Would that work?

Answer 7

In your formula for restoring torque, is this what you are saying:
\[ \tau = \frac{\pi}{2} \frac{R^4}{I} \theta \]
where R = modulus of rigidity and \( I \) is the moment of inertia of the bucket?