EOM of simple pendulum submerged in a fluid

Saptarshi Sarkar
Messages
98
Reaction score
13
Homework Statement
Consider a simple pendulum made of a string of length ℓ and a solid metal bob of radius a (a << ℓ). The bob is completely submerged
in the fluid of viscosity η and the pendulum oscillates back and forth in this fluid. If the density of fluid is ρ0 and that of the metal is ρ
then, assuming slow oscillation, show that the equation of motion for the pendulum is given by (slow oscillation makes linear Stoke’s
law viable, that is resistive force due to viscous fluid is given by 6 π η a v, where v is the instantaneous speed of the bob.)
Relevant Equations
##\ddot \theta(t) = -\left(1-\frac{\rho_0}{\rho}\right)\frac gl sin\theta - \frac{9\eta}{2\rho a^2} \dot \theta##
The question :-

Screenshot_11.png


My attempt :-

1584976267327.jpg


The confusion that I am having is that to get the required form of the equation of motion, I had to approximate ##\theta## to be small to get ##x=l\theta## so that I could get the acceleration and the velocity. But, I had to leave the ##sin(\theta)## in the equation also. Why shouldn't I use ##\theta## instead of ##sin(\theta)## if I have already used that approximation?
 
Physics news on Phys.org
when you write ##g sin \theta## you are evaluating the component of the force along the direction of motion (the tangential component in a circular motion). That direction is not the x direction. So When you write ##\frac {d^2 x} {dt^2}## for the force it's wrong, you should have ##\frac {d^2 s} {dt^2}##, where ##ds## is the displacement in the direction of the motion. It's easy to see that ##ds = l d \theta## and from that follow the equation for the tangential velocity and acceleration. You are not making any small angle approximation here, that comes later when you have to deal with ##sin \theta##
 
Last edited:
  • Informative
Likes Saptarshi Sarkar
dRic2 said:
when you write ##g sin \theta## you are evaluating the component of the force along the direction of motion (the tangential component in a circular motion). That direction is not the x direction. So When you write ##\frac {d^2 x} {dt^2}## for the force it's wrong, you should have ##\frac {d^2 s} {dt^2}##, where ##ds## is the displacement in the direction of the motion. It's easy to see that ##ds = l d \theta## and from that follow the equation for the tangential velocity and acceleration. You are not making any small angle approximation here, that comes later when you have to deal with ##sin \theta##

Thanks !
 
  • Like
Likes dRic2
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
Back
Top