Problems

By pressing a finger down on it, a uniform spherical marble of radius \(a\) is made to slide along a horizontal table top with an initial linear velocity \(v_0\) and an initial {\em backward} angular velocity \(\omega_0\) about the horizontal axis perpendicular to \(v_0\). The frictional force between the marble and the table is constant (independent of speed). For what value of \(v_0/(a\omega_0)\) does the marble

1. slide to a complete stop,
2. come to a stop and then roll back towards its initial position with linear speed \(v_0/7\).

Show Solution

---

Solution:

+ - ↺

If the frictional force is \(F\), then: \begin{align*} L = I\ddot{\theta} && -Fa &= \frac{2}{5}ma^2 \dot{\omega} \\ \text{N2}(\rightarrow) && -F &= m\dot{v} \\ \\ \Rightarrow && \l \frac{2}{5}ma \dot{\omega} - \dot{v} \r &= 0 \\ \Rightarrow && \frac{2}{5}a \omega - v &= c \\ \end{align*}

1. If the ball completely stops, then \(\omega = v = 0 \Rightarrow \frac{2}{5}a \omega_0 - v_0 = 0 \Rightarrow \frac{v_0}{a \omega_0} = \frac25\).
2. If the ball rolls backwards with linear speed \(v_0/7\), \(v = - \frac{v_0}{7}\) and \(a \omega = \frac{v_0}{7}\), \begin{align*} && \frac{2}{5}a \omega_0 - v_0 &= \frac{2}{5} \frac{v_0}{7} + \frac{v_0}{7} \\ && &= \frac{1}{5} v_0 \\ \Rightarrow && \frac{v_0}{a \omega_0} &= \frac{1}{3} \end{align*}