Physics Test Mystery: Paper Flight Through Bernoulli's Theorem

AI Thread Summary
Blowing over a piece of paper creates a high-pressure area in the mouth and releases air, but this does not effectively demonstrate Bernoulli's theorem. The theorem states that increased air velocity above the paper leads to decreased pressure, which should lift the paper. However, the discussion highlights that gravity and static air pressure keep the paper grounded, making it difficult to achieve vertical motion. An alternative demonstration suggests holding two sheets of paper vertically and blowing between them, which may better illustrate the principles of Bernoulli's theorem. This approach emphasizes the complexity of applying Bernoulli's theorem in practical scenarios.
sid_galt
Messages
502
Reaction score
1
I got this question on my physics test

"What will happen if you blow over a piece of paper?"

I answered that with enough blowing, the wind created would make the paper fly off. Apparently according to my teacher, the correct answer was that from Bernouilli's theorem, the air velocity above the paper will increase, the pressure will decrease and the paper will rise up.

But how is that possible? Bernoulli's theorem applies only to a moving fluid whose speed is increased by decreasing the cross sectional area. In blowing, we are creating a high pressure inside the mouth and then releasing the air. The static pressure of the air remains the same all the time. Besides, a paper is held down due to gravity and air pressure. To lift a paper by creating low pressure on top of it would require quite a lot of low pressure plus quite a bit of normal pressure air below the paper.
 
Physics news on Phys.org
sid_galt said:
I got this question on my physics test

"What will happen if you blow over a piece of paper?"

I answered that with enough blowing, the wind created would make the paper fly off. Apparently according to my teacher, the correct answer was that from Bernouilli's theorem, the air velocity above the paper will increase, the pressure will decrease and the paper will rise up.

But how is that possible? Bernoulli's theorem applies only to a moving fluid whose speed is increased by decreasing the cross sectional area. In blowing, we are creating a high pressure inside the mouth and then releasing the air. The static pressure of the air remains the same all the time. Besides, a paper is held down due to gravity and air pressure. To lift a paper by creating low pressure on top of it would require quite a lot of low pressure plus quite a bit of normal pressure air below the paper.

Tried blowing over a piece of paper?
 
yeah. no effect except that the paper gets pushed across the table. But no vertical motion.
 
That's a crummy way to demonstrate Bernoulli, since you can blow away the paper. Try it like this:
Hold a piece (~A4) of paper in each hand vertically next to each other, about 5 cm apart. Hold them at the top with the long side down. Now blow a strong, steady column of air (taut your lips) between the sheets. What will happen according to Bernoulli?
 
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...

Similar threads

Replies
30
Views
4K
Replies
34
Views
3K
Replies
31
Views
4K
Replies
1
Views
2K
Replies
44
Views
10K
Replies
3
Views
2K
Back
Top