Question

4. (20 pts) Curling is clearly the most exciting sport on Earth. Let's pretend that the bottoms of curling stones are perfectly flat, round, and 12 cm in diameter. Let's also assume that the layer of water overlying the ice is 0.5 mm thick. There probably isn't pressure melting, but we will still assume the water is 4°C, at which temperature the dynamic viscosity is 0.0015 N-s/m2. a) What force must be applied to the side of the stone in order for it to ou ду b) What if they wanted to make a Summer Olympics version of curling, by playing on a travel at a constant speed of 1.1 m/s. Report your answer in BOTH newtons and pounds. layer of oil instead of water? At 20°C, SAE 10 oil has a dynamic viscosity of 0.079 N-s/m2. How thick would the oil need to be in order to match the shear stress you just calculated for the winter version of the sport?

Hydrology question

Answer 1

Given data diameter (d) = 12 cm = 0.12 m The layer of water overlying The ice is (t) = 0.5 mm (thick) Dynamic Viscosity (v) = 0.0015 N-s/m^2 Speed (V) = 1.1 m/s Given figure as shown in below. According to Newton's law of Viscosity T alpha dV/dy = > T = mu (dv/dy) \r\n where u = viscosity T = shear stress = F/A dv/dy = Rate of shear deformation F/A = mu (dv/dy) where dV = 1.1 m/s dy = 0.5 mm F = A times u (dv/dy) Therefore Area formula = pi/4 times d^2 F = pi/4 times d^2 times u (dv/dy) here V or u = 0.0015 N-s/m^2 F = pi/4 (0.12)^2 times (0.0015) [1.1/0.5 times 10^-3] = pi/4 (0.0144) times (0.0015) [1.1/5 times 10^-4] = pi/4 times 2.16 times 10^-5 times 2200 \r\n = pi/4 times 0.05 72 = 0.785 times 0.0572 F = 0.0449N and F = 0.00989 Pounds b) here dynamic viscosity