Certainly! Here's a step-by explanation to determine torque required to the circular plate: --- ### Given: - Diameter the plate, \( = 10 \text in} \) - Radius of the plate, \( = \frac{D}{2} =5 \text in} \) - Gap between plates, \( h = .1text{ in} \) - Viscosity of glycerin, \( \eta = 3.13 \times 10^{-2} \text{ lb·s/ft}^2 \) - Rotation speed, \( \omega = 4 \text{ rpm} \) - Shear stress at the edge is negligible - The velocity distribution is linear --- ### Step 1: Convert rotation speed to rad/sec \[ \omega = 4 \text{ rpm} \] \[ \omega_{rad/sec} = \frac{4 \times 2\pi}{60} = \frac{8\pi}{60} = \frac{2\pi}{15} \text{ rad/sec} \] --- ### Step 2: Calculate the velocity at the outer edge \[ v = R \times \omega_{rad/sec} \] Convert \( R \) from inches to feet: \[ R = 5 \text{ in} = \frac{5}{12} \text{ ft} \] \[ v = \frac{5}{12} \times \frac{2\pi}{15} = \frac{5 \times 2\pi}{12 \times 15} = \frac{10\pi}{180} = \frac{\pi}{18} \text{ ft/sec} \] --- ### Step 3: Find the shear stress Since the velocity distribution is linear across the gap: \[ \tau = \eta \times \frac{v}{h} \] Convert \( h \) from inches to feet: \[ h = \frac{.1}{12} \text{ ft} \] \[ \tau = 3.13 \times 10^{-2} \times \frac{\pi/18}{.1/12} \] Simplify denominator: \[ \frac{\pi/18}{.1/12} = \frac{\pi/18}{.00833} \] \[ \Rightarrow \tau = 3.13 \times 10^{-2} \times \frac{\pi/18}{.00833} \] Calculate: \[ \frac{\pi/18}{.00833} = \frac{\pi}{18 \times .00833} \] \[ = \frac{\pi}{.14994} \approx \frac{3.1416}{.14994} \approx 20.96 \] So: \[ \tau \approx 3.13 \times 10^{-2} \times 20.96 \approx .657 \text{ lb/ft}^2 \] --- ### Step 4: Calculate the torque The torque \( T \) is given by integrating shear stress over the surface area: \[ T = \int_{}^{R} \tau \times 2\pi r^2 dr \] Since \( \tau \) varies linearly with \( r \), and shear stress at the edge is primarily responsible: \[ T \approx \tau \times \text{Area} \times R \] \[ T = \tau \times \pi R^2 \times R \] \[ T = \tau \times \pi R^3 \] Convert \( R \) to feet: \[ R = 5/12 \text{ ft} \] Calculate: \[ T = .657 \times \pi \times \left(\frac{5}{12}\right)^3 \] \[ T = .657 \times 3.1416 \times \frac{125}{1728} \] \[ T \approx .657 \times 3.1416 \times .0724 \] \[ T \approx .657 \times .227 \approx .149 \text{ ft·lb} \] --- ### **Final Answer:** \[ \boxed{ \text{Torque} \approx .149 \text{ ft·lb} } \]