How Does a Leaking Cubical Pendulum Affect Its Period?

  • Thread starter Thread starter kashiark
  • Start date Start date
  • Tags Tags
    Pendulum
AI Thread Summary
The discussion focuses on determining the period of a leaking cubical pendulum with a side length of 2a. The relevant equations include the period formula T = 2π√(l/g) and the volume equations for a cube and rectangle. The volume of the pendulum is expressed as v = 4a²h, and the length of the pendulum changes with the new center of mass, leading to l = h/2. By differentiating the volume with respect to time, the relationship between the change in length and the rate of volume change is established. The final expression for the period incorporates the initial length and the rate of leakage, confirming the approach taken is correct.
kashiark
Messages
210
Reaction score
0

Homework Statement


There is a cubical pendulum with a side of 2a filled with water that is leaking at a constant rate. Determine the period of the pendulum in terms of variables.

Homework Equations


T = 2π√(l/g)
v = s³ for a cube
v = s²h for a rectangle

The Attempt at a Solution


v=s²h
v=4a²h
The length varies with the new center of mass which is h/2, so l = h/2.
v=8a²l
differentiate with respect to time and
dv/dt = 8a²dl/dt
dl/dt = 1/(8a²)dv/dt
to avoid the problem from looking too complicated, let's set dv/dt = r
li= initial length
T = 2π√((li-(rt/(8a²)))/g)

I multiplied by time so that we would be left with the change in length, and I subtracted it because dl/dt is negative, and the length should obviously be increasing. Is this right?
 
Physics news on Phys.org
It is correct.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top