RBD CYU 25 - Physics.lt

Physics Solution Q25

Solution to Question 25

Figure: Cross-section of cones A and B showing the neutral point $x_0$ where tangential velocities match.

1. Geometry and Kinematics

Let the total length of the cones be $L$ and the base radius be $R$. We set the origin at the left end ($x=0$). The radii of cones A (top) and B (bottom) at distance $x$ are:

$$r_{\text{A}}(x) = R\left(1 – \frac{x}{L}\right)$$ $$r_{\text{B}}(x) = R\left(\frac{x}{L}\right)$$

The tangential velocities at the interface are:

$$v_{\text{A}} = \omega_{\text{A}} r_{\text{A}}(x)$$ $$v_{\text{B}} = \omega_{\text{B}} r_{\text{B}}(x)$$

2. Finding the Neutral Point ($x_0$)

Friction accelerates cone B where $v_{\text{A}} > v_{\text{B}}$ and retards it where $v_{\text{A}} < v_{\text{B}}$. In the steady state, there is a neutral point $x_0$ where no slipping occurs ($v_{\text{A}} = v_{\text{B}}$):

$$\omega_{\text{A}} R \left(1 – \frac{x_0}{L}\right) = \omega_{\text{B}} R \left(\frac{x_0}{L}\right)$$ $$\Rightarrow \omega_{\text{A}} – \omega_{\text{A}}\frac{x_0}{L} = \omega_{\text{B}}\frac{x_0}{L}$$ $$\Rightarrow \frac{x_0}{L} = \frac{\omega_{\text{A}}}{\omega_{\text{A}} + \omega_{\text{B}}} \quad \dots (1)$$

3. Torque Equilibrium

Let the normal force per unit length be $\lambda$. The torque on a small element $dx$ of cone B is $d\tau = r_{\text{B}}(x) (\mu \lambda dx)$. For steady angular velocity, the net torque on B must be zero. This means the accelerating torque (from $0$ to $x_0$) equals the retarding torque (from $x_0$ to $L$):

$$\int_{0}^{x_0} r_{\text{B}}(x) \, dx = \int_{x_0}^{L} r_{\text{B}}(x) \, dx$$ $$\int_{0}^{x_0} \frac{R}{L}x \, dx = \int_{x_0}^{L} \frac{R}{L}x \, dx$$

Canceling constants and integrating:

$$\left[ \frac{x^2}{2} \right]_{0}^{x_0} = \left[ \frac{x^2}{2} \right]_{x_0}^{L}$$ $$x_0^2 = L^2 – x_0^2 \implies 2x_0^2 = L^2 \implies x_0 = \frac{L}{\sqrt{2}} \quad \dots (2)$$

4. Final Calculation

Substitute result (2) into equation (1):

$$\frac{1}{\sqrt{2}} = \frac{\omega_{\text{A}}}{\omega_{\text{A}} + \omega_{\text{B}}}$$ $$\omega_{\text{A}} + \omega_{\text{B}} = \sqrt{2}\omega_{\text{A}}$$ $$\omega_{\text{B}} = (\sqrt{2} – 1)\omega_{\text{A}}$$