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A cylinder rapidly spinning about horizontal axis was dropped from height h=5 meter?
and landed at distance d = 0.5 meter from the the point directly below its initial position.
Cylinder:
mass m = 10 kg
length L = 1 meter
radius r = 10 cm
Environment:
acceleration of gravity g = 10 m/s²
density of air ρ = 1kg/m³
How fast was the cylinder spinning?
2 Answers
- ?Lv 51 decade agoFavorite Answer
Neglect the air resistance and boundary layer slip.
Due to the Magnus effect, the cylinder accelerates in the direction normal to its velocity V. The corresponding acceleration is proportional to V: a = k V. Motion along the horizontal axis x is described by
d V_x/ dt = k V cos θ = k d z/ dt.
Here, θ is the angle between V and the downwards direction z. The Magnus force is normal to V and it does not change the conservation law. Setting z=0 at the initial height,
V_x^2 + V_z^2 = 2 g z.
Using above equations one has
dx / dz = V_x/ V_z = k z / √ ( 2 g z - k^2 z^2).
The second term under the radical is about (d/h)^2 = 0.01 smaller than the first term, and it can be neglected. This gives
d = k ∫ √ ( z / 2 g) d z = √2 k h^(3/2)/ (3 √g) ,
k = 3 √ (g h/ 2) d / h^2.
The lifting force is due to the pressure difference, which at the opposite points of the cylinder is equal to
δ p_max = (ρ/2) [ (ω r + V)^2 - (ω r - V)^2] = 2 ρ ω r V.
δ p is equal to δ p_max cos φ where φ is the angle around the cylinder. Another cosine dependence appears due to projection of the force on the x-axis. The force per unit length is ∫ r delta; p_max cos^2 d φ = 2 π ρ ω r^2 V. Hence,
k = 2 π ρ ω r^2 L / m,
ω = (3 / π) √ (g h/ 8) (m / ρ) d / (h^2 r^2 L) =
= (3 / π) √ (10* 5/ 8)* (10 / 1) * 0.5/ (5^2 * 0.1^2*1) = 150 /π.
The cylinder makes about 7.6 rotations per second.
With the boundary layer slip, one possibly needs a couple of times faster rotation to get the same effect.
- Anonymous4 years ago
No it gained't leap extra useful situations yet lots much less situations . with the aid of fact the bouncing is a reaction the ball does after it hits the floor and the potential of this reaction equals the potential (potential) of the action . And the action right this is the load of the ball and so long with the aid of fact the gravity of the moon is sixteen of the gravity on the earth then the action would be sixteen of the action on the earth and likewise the reaction (this is in many cases yet there are diverse matters that effect like the situation you're doing the journey is it flat or not so the lost potential would be lots much less )