Let's address each part of the problem for a ball of mass \( m \) rolling inside the paraboloid \( z = x^2 + y^2 \) under gravity. --- ### (a) **Newton’s Equations** The ball is constrained to the surface \( z = x^2 + y^2 \). The only forces acting are gravity and the normal reaction from the surface. The equations of motion are: 1. **Force components:** - Gravity: \( \vec{F}_g = -mg \hat{z} \) - Normal force: \( \vec{N} \) (unknown, but perpendicular to the surface) 2. **Constraint equation:** \[ z - x^2 - y^2 = 0 \] 3. **Newton’s second law (with constraint):** \[ m\ddot{\vec{r}} = -mg\hat{z} + N\hat{n} \] where \( \hat{n} \) is the unit normal to the surface. 4. **Unit normal:** \[ \hat{n} = \frac{\nabla f}{|\nabla f|}, \quad f(x, y, z) = z - x^2 - y^2 \] \[ \nabla f = (-2x, -2y, 1) \] \[ |\nabla f| = \sqrt{4x^2 + 4y^2 + 1} \] So, \[ \hat{n} = \frac{(-2x, -2y, 1)}{\sqrt{4x^2 + 4y^2 + 1}} \] 5. **For \( x \) and \( y \):** Project the force balance onto the tangent plane (eliminate the normal force component). The tangent plane is spanned by \( \hat{x} \) and \( \hat{y} \), so write Newton’s equation for \( x \) and \( y \): \[ m\ddot{x} = -N\frac{2x}{\sqrt{4x^2 + 4y^2 + 1}} \] \[ m\ddot{y} = -N\frac{2y}{\sqrt{4x^2 + 4y^2 + 1}} \] Use the constraint to relate \( \ddot{z} \) to \( \ddot{x} \) and \( \ddot{y} \): \[ z = x^2 + y^2 \implies \dot{z} = 2x\dot{x} + 2y\dot{y} \] \[ \ddot{z} = 2\dot{x}^2 + 2x\ddot{x} + 2\dot{y}^2 + 2y\ddot{y} \] The full equations can be derived by substituting for \( N \) using the constraint and the full force balance in \( z \). --- ### (b) **Principle of Virtual Power** The virtual power principle states that for each virtual velocity \( (\delta \dot{x}, \delta \dot{y}) \) consistent with the constraint, \[ \sum \vec{F}_i \cdot \delta \vec{v}_i = 0 \] where \( \vec{F}_i \) are the applied forces, and \( \delta \vec{v}_i \) are admissible virtual velocities. For our case, the admissible virtual displacements \( (\delta x, \delta y, \delta z) \) must satisfy the constraint: \[ \delta z = 2x\delta x + 2y\delta y \] The virtual work (power) done by gravity: \[ \delta W = -mg \delta z = -mg (2x \delta x + 2y \delta y) \] So, the virtual power equation (for kinetic energy change) is: \[ m(\ddot{x} \delta x + \ddot{y} \delta y + \ddot{z} \delta z) = -mg \delta z \] But with the constraint linking \( \delta z \) to \( \delta x, \delta y \), you substitute and equate coefficients to find the equations of motion. --- ### (c) **Lagrange’s Equations** The kinetic energy (assuming the ball is sliding, not rolling): \[ T = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2) \] but \( \dot{z} = 2x\dot{x} + 2y\dot{y} \), so: \[ T = \frac{1}{2}m\left[\dot{x}^2 + \dot{y}^2 + (2x\dot{x} + 2y\dot{y})^2\right] \] \[ = \frac{1}{2}m\left[\dot{x}^2 + \dot{y}^2 + 4x^2\dot{x}^2 + 8xy\dot{x}\dot{y} + 4y^2\dot{y}^2\right] \] \[ = \frac{1}{2}m\left[(1+4x^2)\dot{x}^2 + (1+4y^2)\dot{y}^2 + 8xy\dot{x}\dot{y}\right] \] The potential energy: \[ V = mgz = mg(x^2 + y^2) \] The Lagrangian: \[ L = T - V \] The Euler-Lagrange equations: \[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_i} \right) - \frac{\partial L}{\partial q_i} = 0, \quad q_i = x, y \] Apply for \( x \) and \( y \): \[ \frac{d}{dt} \left( m[(1+4x^2)\dot{x} + 4xy\dot{y}] \right) - 2mgx = 0 \] \[ \frac{d}{dt} \left( m[(1+4y^2)\dot{y} + 4xy\dot{x}] \right) - 2mgy = 0 \] --- #### **Summary of Equations of Motion:** - **Newton:** Use force balance and eliminate normal force using the constraint. - **Virtual Power:** Project dynamics onto virtual displacements compatible with the constraint. - **Lagrange:** Use the Lagrangian above and derive the Euler-Lagrange equations for \( x \) and \( y \). If you need a more detailed derivation for any part, let me know!