1. **Problem Statement:** A particle of mass $2m$ is projected horizontally with velocity $u$ along the smooth inner surface of a cone with semi-vertical angle $\alpha$, vertex downwards, and axis vertically upwards. Initially, the particle is at height $\frac{u^2}{g}$ above the vertex. The particle's position at time $t$ is given by cylindrical coordinates $(r, \theta, z)$. We need to: (a) Find the height from the vertex to the particle when $\dot{z} = 0$. (b) Obtain an expression for $\dot{z}$ at any time. (c) Determine whether the particle moves up or down. --- 2. **Geometry and Constraints:** The cone's surface satisfies the relation between $r$ and $z$: $$r = z \tan \alpha$$ Since the particle moves on the cone surface, this relation holds at all times. --- 3. **Initial Conditions:** Initial height: $$z_0 = \frac{u^2}{g}$$ Initial velocity is horizontal with magnitude $u$, so initial vertical velocity $\dot{z}(0) = 0$. --- 4. **Kinematics and Dynamics:** The particle moves on the cone surface, so its position is constrained. The velocity components relate as: $$r = z \tan \alpha \implies \dot{r} = \dot{z} \tan \alpha$$ The particle is projected horizontally, so initial vertical velocity $\dot{z}(0) = 0$. --- 5. **Energy Conservation:** The particle moves without friction on a smooth surface, so mechanical energy is conserved. Kinetic energy: $$T = \frac{1}{2} (2m) (\dot{r}^2 + r^2 \dot{\theta}^2 + \dot{z}^2)$$ Potential energy: $$V = 2mgz$$ Using the constraint $r = z \tan \alpha$, substitute $r$ and $\dot{r}$: $$\dot{r} = \dot{z} \tan \alpha$$ So kinetic energy becomes: $$T = m \left( \dot{z}^2 \tan^2 \alpha + z^2 \tan^2 \alpha \dot{\theta}^2 + \dot{z}^2 \right) = m \left( \dot{z}^2 (1 + \tan^2 \alpha) + z^2 \tan^2 \alpha \dot{\theta}^2 \right)$$ Recall $1 + \tan^2 \alpha = \sec^2 \alpha$: $$T = m \left( \dot{z}^2 \sec^2 \alpha + z^2 \tan^2 \alpha \dot{\theta}^2 \right)$$ --- 6. **Angular Momentum Conservation:** No torque about the vertical axis implies angular momentum about $z$-axis is conserved: $$L_z = 2m r^2 \dot{\theta} = \text{constant}$$ Substitute $r = z \tan \alpha$: $$L_z = 2m (z^2 \tan^2 \alpha) \dot{\theta} = h$$ Solve for $\dot{\theta}$: $$\dot{\theta} = \frac{h}{2m z^2 \tan^2 \alpha}$$ --- 7. **Energy Expression:** Total energy $E$ is constant: $$E = T + V = m \dot{z}^2 \sec^2 \alpha + m z^2 \tan^2 \alpha \dot{\theta}^2 + 2mgz$$ Substitute $\dot{\theta}$: $$E = m \dot{z}^2 \sec^2 \alpha + m z^2 \tan^2 \alpha \left( \frac{h}{2m z^2 \tan^2 \alpha} \right)^2 + 2mgz$$ Simplify second term: $$= m \dot{z}^2 \sec^2 \alpha + m z^2 \tan^2 \alpha \frac{h^2}{4 m^2 z^4 \tan^4 \alpha} + 2mgz$$ $$= m \dot{z}^2 \sec^2 \alpha + \frac{h^2}{4 m z^2 \tan^2 \alpha} + 2mgz$$ --- 8. **At $\dot{z} = 0$ (part a):** Set $\dot{z} = 0$ in energy equation: $$E = \frac{h^2}{4 m z^2 \tan^2 \alpha} + 2mgz$$ Initially, at $z_0 = \frac{u^2}{g}$, $\dot{z} = 0$, so energy is: $$E = m \cdot 0 \cdot \sec^2 \alpha + m z_0^2 \tan^2 \alpha \dot{\theta}_0^2 + 2mg z_0$$ Initial angular velocity $\dot{\theta}_0$ is found from initial horizontal velocity $u$: Horizontal velocity is along the cone surface, so: $$u^2 = \dot{r}^2 + r^2 \dot{\theta}^2$$ Since initial vertical velocity $\dot{z} = 0$, $\dot{r} = 0$, so: $$u = r \dot{\theta}_0 = z_0 \tan \alpha \dot{\theta}_0$$ Solve for $\dot{\theta}_0$: $$\dot{\theta}_0 = \frac{u}{z_0 \tan \alpha}$$ Substitute into initial energy: $$E = m z_0^2 \tan^2 \alpha \left( \frac{u}{z_0 \tan \alpha} \right)^2 + 2mg z_0 = m u^2 + 2mg z_0$$ Recall $z_0 = \frac{u^2}{g}$: $$E = m u^2 + 2mg \frac{u^2}{g} = m u^2 + 2 m u^2 = 3 m u^2$$ --- 9. **Find $z$ when $\dot{z} = 0$:** From step 8: $$E = \frac{h^2}{4 m z^2 \tan^2 \alpha} + 2mg z = 3 m u^2$$ Recall angular momentum $h$: $$h = 2m r^2 \dot{\theta} = 2m z_0^2 \tan^2 \alpha \dot{\theta}_0 = 2m z_0^2 \tan^2 \alpha \frac{u}{z_0 \tan \alpha} = 2m z_0 u \tan \alpha$$ Simplify: $$h = 2m u z_0 \tan \alpha$$ Substitute $h$ into energy equation: $$3 m u^2 = \frac{(2m u z_0 \tan \alpha)^2}{4 m z^2 \tan^2 \alpha} + 2 m g z$$ Simplify numerator: $$= \frac{4 m^2 u^2 z_0^2 \tan^2 \alpha}{4 m z^2 \tan^2 \alpha} + 2 m g z = \frac{m u^2 z_0^2}{z^2} + 2 m g z$$ Divide both sides by $m$: $$3 u^2 = \frac{u^2 z_0^2}{z^2} + 2 g z$$ Multiply both sides by $z^2$: $$3 u^2 z^2 = u^2 z_0^2 + 2 g z^3$$ Rearranged: $$2 g z^3 - 3 u^2 z^2 + u^2 z_0^2 = 0$$ Recall $z_0 = \frac{u^2}{g}$: $$2 g z^3 - 3 u^2 z^2 + u^2 \left( \frac{u^4}{g^2} \right) = 0$$ Simplify: $$2 g z^3 - 3 u^2 z^2 + \frac{u^6}{g^2} = 0$$ This cubic equation in $z$ gives the height when $\dot{z} = 0$. --- 10. **Expression for $\dot{z}$ (part b):** From energy conservation: $$E = m \dot{z}^2 \sec^2 \alpha + \frac{h^2}{4 m z^2 \tan^2 \alpha} + 2 m g z$$ Solve for $\dot{z}^2$: $$\dot{z}^2 = \frac{1}{m \sec^2 \alpha} \left( E - \frac{h^2}{4 m z^2 \tan^2 \alpha} - 2 m g z \right)$$ Simplify: $$\dot{z} = \pm \sqrt{ \frac{1}{m \sec^2 \alpha} \left( E - \frac{h^2}{4 m z^2 \tan^2 \alpha} - 2 m g z \right) }$$ Substitute $E = 3 m u^2$ and $h = 2 m u z_0 \tan \alpha$: $$\dot{z} = \pm \sqrt{ \frac{1}{m \sec^2 \alpha} \left( 3 m u^2 - \frac{4 m^2 u^2 z_0^2 \tan^2 \alpha}{4 m z^2 \tan^2 \alpha} - 2 m g z \right) } = \pm \sqrt{ \frac{1}{m \sec^2 \alpha} \left( 3 m u^2 - \frac{m u^2 z_0^2}{z^2} - 2 m g z \right) }$$ Cancel $m$: $$\dot{z} = \pm \sqrt{ \frac{1}{\sec^2 \alpha} \left( 3 u^2 - \frac{u^2 z_0^2}{z^2} - 2 g z \right) } = \pm \cos \alpha \sqrt{ 3 u^2 - \frac{u^2 z_0^2}{z^2} - 2 g z }$$ --- 11. **Direction of motion (part c):** Initially, $\dot{z} = 0$ at $z = z_0$. To determine if the particle moves up or down, consider the acceleration $\ddot{z}$ or the sign of $\dot{z}$ immediately after projection. From the energy expression, the particle will move towards decreasing potential energy, so it will move downwards from the initial height because gravity acts downward and the particle has no initial vertical velocity. Hence, the particle moves down the cone surface. --- **Final answers:** (a) Height $z$ when $\dot{z} = 0$ satisfies: $$2 g z^3 - 3 u^2 z^2 + u^2 z_0^2 = 0$$ with $z_0 = \frac{u^2}{g}$. (b) Expression for $\dot{z}$: $$\dot{z} = \pm \cos \alpha \sqrt{ 3 u^2 - \frac{u^2 z_0^2}{z^2} - 2 g z }$$ (c) The particle moves downwards from the initial height.