In a dramatic lecture demonstration, a physics professor blows hard across the top of a copper penny that is at rest on a level desk. By doing this at the right speed, he can get the penny to accelerate vertically, into the airstream, and then deflect it into a tray.
Assuming the diameter of a penny is 1.70 cm and its mass is 2.80 g, what is the minimum air speed needed to lift the penny off the tabletop? Assume the air under the penny remains at rest.
Require that the Bernoulli-effect pressure difference above and below the penny balance the weight.
(1/2)(rho)V^2*pi*D^2/4 = M g
M is the penny's mass and rho is the density of air, about 1.3 kg/m^3.
Solve for V.
V = sqrt[8*M*g/(pi*D^2*rho)]
= 13.6 m/s
(about 30.5 mph)
To determine the minimum air speed needed to lift the penny off the tabletop, we need to consider the forces acting on the penny.
In this case, there are two main forces: the force of gravity pulling the penny down and the force of the air pushing the penny up.
To lift the penny off the tabletop, the force exerted by the air pushing the penny up needs to be greater than the force of gravity pulling it down.
We can use Bernoulli's principle to calculate the minimum air speed required. Bernoulli's principle states that as the speed of a fluid (in this case, air) increases, the pressure of the fluid decreases.
Here's the step-by-step process to determine the minimum air speed:
1. Calculate the weight of the penny using the formula: weight = mass × acceleration due to gravity. The acceleration due to gravity is approximately 9.8 m/s². Convert the mass of the penny from grams to kilograms.
weight = (2.80 g) × (9.8 m/s²) = 0.002744 N
2. The force exerted by the air pushing the penny up is equal to the pressure difference multiplied by the area over which the pressure difference acts. In this case, the pressure difference is the difference between atmospheric pressure and the pressure on top of the penny, and the area is the cross-sectional area of the penny.
Area = π × (diameter/2)²
Area = π × (1.70 cm/2)² = 0.002262 m²
3. The pressure difference can be calculated using Bernoulli's principle. At the minimum air speed, the pressure on top of the penny is atmospheric pressure (since the penny is just about to lift off), which is approximately 101,325 Pa.
P1 - P2 = ½ × ρ × v²
P1 - 101,325 Pa = 0.5 × (density of air) × (velocity of air)²
Since the air under the penny remains at rest, the density of air is the density at rest, which is approximately 1.225 kg/m³.
4. Rearrange the equation from step 3 to solve for the velocity of air:
(velocity of air)² = (P1 - 101,325 Pa) / (0.5 × 1.225 kg/m³)
velocity of air = √[(P1 - 101,325 Pa) / (0.5 × 1.225 kg/m³)]
5. Substitute the value of P1 into the equation from step 4. Since P1 is atmospheric pressure, we get:
velocity of air = √[(101,325 Pa - 101,325 Pa) / (0.5 × 1.225 kg/m³)]
velocity of air = √[0 / (0.5 × 1.225 kg/m³)] = 0 m/s
Therefore, the minimum air speed needed to lift the penny off the tabletop is 0 m/s.