What is the Optimum Height for Drinking a Milkshake with a Vertical Straw?

AI Thread Summary
The discussion centers on the physics of drinking a milkshake through a vertical straw, specifically examining the height at which the straw must be positioned for effective drinking. It is established that the pressure head generated by human lungs (3000 Pa) is insufficient to overcome the height of the milkshake in the straw, preventing drinking. For optimal height, the straw must be positioned just below the pressure head limit to allow the milkshake to flow. Bernoulli's equation is referenced as a potential method for calculations, but a simpler pressure difference approach is deemed sufficient. Ultimately, the participant successfully resolves the problem after clarification on the concepts involved.
Adriano Roma
Messages
2
Reaction score
0
Member advised to use the formatting template for questions posted to the homework forums
You and a friend of yours went for a drink of chocolate milkshake(ρ=1200 kg/m3). The waiter brings your drink in two glasses 200 mm tall with a straw 8 mm in diameter and 300 mm long. Given that human lungs capacity can develop approximately 3000 Pa of vacuum pressure and assuming that the straw is barely inserted into the milkshake:
(a) Verify that you won't be able to drink the milkshake through the vertical straw.
(b) Determine the optimum height of the vertical straw above the milkshake that allows the drink to just flow to your mouth.

As far as I understood, for point (a) I think I need to verify that the pressure head corresponding to the pressure difference between the bottom and the top of the straw is shorter than the straw and for point (b) I would say that the hight of the straw above the drink has to be slightly less than the pressure head corresponding to the pressure difference.

I think I need to apply Bernoulli's equation but don't know how to proceed!

Could you please give me some hints? I attached an image which illustrates the problem.

Thank you in advance.
 

Attachments

  • Prob.jpg
    Prob.jpg
    10 KB · Views: 512
Physics news on Phys.org
Hello Adriano, benvenuto :welcome:

Bernoulli's equation is an energy balance. In this case a pressure difference calculation is sufficient. For a, and for b too !

PS nice illustration !
 
  • Like
Likes Adriano Roma
Thank you very much, I managed to solve it. I just didn't understand Bernoulli's equation!
Regards
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top