# Spinning ball bounces off ground, following curved trajectory

by azure kitsune
Tags: ball, bounces, curved, ground, spinning, trajectory
 P: 65 1. The problem statement, all variables and given/known data You have a solid, homogenous ball with radius R. Before falling to the floor, its center of mass is at rest, but it is spinning with angular velocity $$\omega_0$$ about a horizontal axis through its center. The lowest point of the ball is at a height h above the floor. When released, the ball falls under the influence of gravity, and rebounds to a new height such that its highest point is $$\alpha h$$ above the floor. The deformation of the ball and the floor due to the impact can be negligible; the impact time, though, is finite. The mass of the ball is m, and the coefficient of kinetic friction between the ball and the floor is $$\mu_k$$. Ignore air resistance. For the situation where the ball is slipping throughout the impact, find the horizontal distance traveled in flight between the first and second impacts. 2. Relevant equations Conservation of energy, momentum Equation for friction. 3. The attempt at a solution I have no idea how to start this problem. In between collisions, the angular velocity of the ball remains constant, right? That means if I can find the translational velocity (magnitude and direction) immediately after collision, I will be able to answer this problem. I was thinking of using conservation of energy. Since there is no deformation, this means all the energy lost in the collision is due to friction right? But this does not help me find at what angle the ball bounces off the ground, which I need.
 P: 65 Is this correct? Let $$v_x$$ and $$v_y$$ be the x and y components of the velocity of the ball immediately after impact. It's easy to calculate that $$v_y = \sqrt{2g\alpha h}$$. Let $$\Delta t$$ be the time of the impact. During this time, we have $$\Delta v_y = \sqrt{2gh} + \sqrt{2g\alpha h} = \sqrt{2gh}(\sqrt{\alpha}+1)$$ The average normal force exerted during this time is given by: $$\Sigma F_y = N - mg = m \frac{\sqrt{2gh}(\sqrt{\alpha}+1)}{\Delta t}$$ So the average friction (net force in x-direction) during this time is $$f = m \frac{\Delta v_x}{\Delta t} = \mu N$$ So $$v_x = \mu (g\Delta t + \sqrt{2gh}(\sqrt{\alpha}+1))$$ Thus, we have: $$v_y = \sqrt{2g\alpha h}$$ $$v_x = \mu (g\Delta t + \sqrt{2gh}(\sqrt{\alpha}+1))$$ which is enough to solve the problem. Is it okay to assume that $$g\Delta t$$ is much smaller than $$\sqrt{2gh}(\sqrt{\alpha}+1)$$, so we can say $$v_x = \mu\sqrt{2gh}(\sqrt{\alpha}+1)$$ ?