Let's solve the problem step by step. --- ## **Given Data** - **Angular speed,** \( \omega = 42 \) rad/s - **Viscosity,** \( \mu = .1 \) Pa·s - **Film thickness,** \( t = .025 \) cm = \( .00025 \) m - **Base radius of cone,** \( R = 10 \) cm = \( .1 \) m - **Length of cone,** \( h = 30 \) cm = \( .3 \) m --- ## **Step 1: Velocity Gradient and Shear Stress** The oil film thickness \( t \) is very small. Assuming the cone rotates and the oil film causes viscous drag: **Velocity at distance \( r \) from the axis:** \[ v = \omega r \] **Velocity gradient:** \[ \frac{dv}{dy} \approx \frac{\omega r}{t} \] **Shear stress:** \[ \tau = \mu \frac{dv}{dy} = \mu \frac{\omega r}{t} \] --- ## **Step 2: Area Element** Take a thin ring at distance \( r \) from the axis, width \( dr \): - **Slant height:** \( h \) - **Corresponding arc length for a full revolution:** \( 2\pi r \) **Area of ring on slant surface:** \[ dA = 2\pi r \cdot dl \] where \( dl \) is the width along the slant. **From the cone geometry:** \[ \frac{r}{R} = \frac{l}{h} \implies dl = \frac{h}{R} dr \] So: \[ dA = 2\pi r \left(\frac{h}{R}\right) dr \] --- ## **Step 3: Differential Force and Torque** **Differential force:** \[ dF = \tau \cdot dA = \mu \frac{\omega r}{t} \cdot 2\pi r \frac{h}{R} dr = 2\pi \mu \frac{\omega h}{t R} r^2 dr \] **Differential torque:** \[ dT = dF \cdot r = 2\pi \mu \frac{\omega h}{t R} r^3 dr \] --- ## **Step 4: Total Torque** Integrate from \( r = \) to \( r = R \): \[ T = \int_^R 2\pi \mu \frac{\omega h}{t R} r^3 dr \] \[ T = 2\pi \mu \frac{\omega h}{t R} \int_^R r^3 dr \] \[ \int_^R r^3 dr = \frac{R^4}{4} \] So, \[ T = 2\pi \mu \frac{\omega h}{t R} \cdot \frac{R^4}{4} = \frac{\pi}{2} \frac{\mu \omega h R^3}{t} \] --- ## **Step 5: Plug in the Values** - \( \mu = .1 \) Pa·s - \( \omega = 42 \) rad/s - \( h = .3 \) m - \( R = .1 \) m - \( t = .00025 \) m \[ T = \frac{\pi}{2} \cdot \frac{.1 \cdot 42 \cdot .3 \cdot (.1)^3}{.00025} \] First, calculate \( (.1)^3 = .001 \). \[ T = \frac{\pi}{2} \cdot \frac{.1 \cdot 42 \cdot .3 \cdot .001}{.00025} \] \[ = \frac{\pi}{2} \cdot \frac{.00126}{.00025} \] \[ = \frac{\pi}{2} \cdot 5.04 \] \[ = 1.57 \cdot 5.04 \approx 7.92 \text{ Nm} \] --- ## **Final Answer** \[ \boxed{T \approx 7.92 \ \text{Nm}} \] **This is the torque required to maintain the motion.**