Physics

Physics · mechanics

Problem

Spinning Cylinder Adithya has a solid cylinder of mass M=10 kgM=10 \mathrm{~kg}, radius R=0.08 mR=0.08 \mathrm{~m}, and height H=0.20 mH=0.20 \mathrm{~m}. He is running a test in a chamber on Earth over a distance of d=200 md=200 \mathrm{~m} as shown below. Assume that the physical length of the chamber is much greater than dd (i.e. the chamber extends far to the left and right of the testing area). The chamber is filled with an ideal fluid with uniform density ρ=700 kg/m3\rho=700 \mathrm{~kg} / \mathrm{m}^{3}. Adithya's cylinder is launched with linear velocity v=10 m/sv=10 \mathrm{~m} / \mathrm{s} and spins counterclockwise with angular velocity ω\omega. Adithya notices that the cylinder continues on a horizontal path until the end of the chamber. Find the angular velocity ω\omega. Do not neglect forces due to fluid pressure differences. Note that the diagram presents a side view of the chamber (i.e. gravity is oriented downwards with respect to the diagram). <image_1> Assume the following about the setup and the ideal fluid: - fluid flow is steady in the frame of the center of mass of the cylinder - the ideal fluid is incompressible, irrotational, and has zero viscosity - the angular velocity of the cylinder is approximately constant during its subsequent motion Hint: For a uniform cylinder of radius RR rotating counterclockwise at angular velocity ω\omega situated in an ideal fluid with flow velocity uu to the right far away from the cylinder, the velocity potential Φ\Phi is given by Φ(r,θ)=urcos⁡θ+uR2rcos⁡θ+Γθ2π\Phi(r, \theta)=u r \cos \theta+u \frac{R^{2}}{r} \cos \theta+\frac{\Gamma \theta}{2 \pi} where (r,θ)(r, \theta) is the polar coordinate system with origin at the center of the cylinder. Γ\Gamma is the circulation and is equal to 2πR2ω2 \pi R^{2} \omega. The fluid velocity is given by v=∇Φ=∂Φ∂rr^+1r∂Φ∂θθ^\mathbf{v}=\nabla \Phi=\frac{\partial \Phi}{\partial r} \hat{\mathbf{r}}+\frac{1}{r} \frac{\partial \Phi}{\partial \theta} \hat{\theta} Given constants: - Proton mass, mp=1.67⋅10−27 kgm_{p}=1.67 \cdot 10^{-27} \mathrm{~kg} - Neutron mass, mn=1.67⋅10−27 kgm_{n}=1.67 \cdot 10^{-27} \mathrm{~kg} - Electron mass, me=9.11⋅10−31 kgm_{e}=9.11 \cdot 10^{-31} \mathrm{~kg} - Avogadro's constant, N0=6.02⋅1023 mol−1N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1} - Universal gas constant, R=8.31 J/(mol⋅K)R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) - Boltzmann's constant, kB=1.38⋅10−23 J/Kk_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K} - Electron charge magnitude, e=1.60⋅10−19Ce=1.60 \cdot 10^{-19} \mathrm{C} - 1 electron volt, 1eV=1.60⋅10−19 J1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J} - Speed of light, c=3.00⋅108 m/sc=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s} - Universal Gravitational constant, G=6.67⋅10−11( N⋅m2)/kg2G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} - Solar Mass M⊙=1.988⋅1030 kgM_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} - Acceleration due to gravity, g=9.8 m/s2g=9.8 \mathrm{~m} / \mathrm{s}^{2} - 1 unified atomic mass unit, 1u=1.66⋅10−27 kg=931MeV/c21 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} - Planck's constant, h=6.63⋅10−34 J⋅s=4.41⋅10−15eV⋅sh=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} - Permittivity of free space, ϵ0=8.85⋅10−12C2/(N⋅m2)\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) - Coulomb's law constant, k=14πϵ0=8.99⋅109( N⋅m2)/C2k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} - Permeability of free space, μ0=4π⋅10−7 T⋅m/A\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} - Magnetic constant, μ04π=1⋅10−7( T⋅m)/A\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} - 1 atmospheric pressure, 1 atm=1.01⋅105 N/m2=1.01⋅105 Pa1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} - Wien's displacement constant, b=2.9b=2.9. 10−3 m⋅K10^{-3} \mathrm{~m} \cdot \mathrm{K} - Stefan-Boltzmann constant, σ=5.67⋅10−8 W/m2/K4\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
Intuitive explanation

This problem combines buoyancy, gravity, and the Magnus effect—a spinning object moving through a fluid experiences a lift force perpendicular to its motion. The key insight is that the cylinder travels horizontally, meaning the net vertical force must be zero: the upward Magnus force (generated by the spinning motion in the fluid flow) must exactly balance the downward gravitational force minus buoyancy. Use the velocity potential and Bernoulli's equation to find the pressure distribution around the spinning cylinder, which gives you the Magnus force, then set it equal to the net weight to solve for ω\omega.

Intuitive explanation

This problem combines buoyancy, gravity, and the Magnus effect—a spinning object moving through a fluid experiences a lift force perpendicular to its motion. The key insight is that the cylinder travels horizontally, meaning the net vertical force must be zero: the upward Magnus force (generated by the spinning motion in the fluid flow) must exactly balance the downward gravitational force minus buoyancy. Use the velocity potential and Bernoulli's equation to find the pressure distribution around the spinning cylinder, which gives you the Magnus force, then set it equal to the net weight to solve for ω\omega.

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