Source: Physics Competition - Mechanics
Problem Sets:
Problem
A light rigid rod of length $L$ has two small balls, A and B, fixed at its ends. Their masses are $m_A = M$ and $m_B = 2M$, respectively. The system is placed on a smooth horizontal plane. At a certain instant, the speeds of the two balls are $v_A = v$ and $v_B = 2v$.
The possible tension values are $\frac{2Mv^2}{3L} \le T \le \frac{6Mv^2}{L}$.
The motion of each mass can be decomposed into components parallel ($v_{||}$) and perpendicular ($v_{\perp}$) to the rod. The rigidity of the rod requires that the parallel velocity components are equal, $v_{A||} = v_{B||} = v_{||}$.
From the given speeds, we have:
$$v_A^2 = v_{||}^2 + v_{A\perp}^2 = v^2$$ $$v_B^2 = v_{||}^2 + v_{B\perp}^2 = (2v)^2 = 4v^2$$Subtracting the first equation from the second gives a relationship between the perpendicular components:
$$v_{B\perp}^2 - v_{A\perp}^2 = 3v^2$$The tension $T$ in the rod provides the centripetal force for the system's rotation about its center of mass (CM). The CM is located at a distance $r_A = \frac{m_B}{m_A+m_B}L = \frac{2L}{3}$ from mass A, and $r_B = \frac{m_A}{m_A+m_B}L = \frac{L}{3}$ from mass B.
The angular velocity of the rod is $\omega = \frac{|v_{A\perp} - v_{B\perp}|}{L}$. The tension can be found from either mass:
$$T = m_A r_A \omega^2 = M \left(\frac{2L}{3}\right) \left(\frac{v_{A\perp} - v_{B\perp}}{L}\right)^2 = \frac{2M}{3L}(v_{A\perp} - v_{B\perp})^2$$The problem asks for possible values of tension. The extreme values occur when the kinetic energy in rotation is maximized, which means the translational velocity component along the rod is zero ($v_{||}=0$). If $v_{||}=0$, then from the speed equations, $|v_{A\perp}| = v$ and $|v_{B\perp}| = 2v$. This is consistent with the relation $v_{B\perp}^2 - v_{A\perp}^2 = 3v^2$, since $(2v)^2 - v^2 = 3v^2$.
There are two possible scenarios for this purely rotational motion:
- The perpendicular velocities are in the same direction ($v_{A\perp}$ and $v_{B\perp}$ have the same sign). This corresponds to $\vec{v}_A$ and $\vec{v}_B$ being parallel.
- The perpendicular velocities are in opposite directions ($v_{A\perp}$ and $v_{B\perp}$ have opposite signs). This corresponds to $\vec{v}_A$ and $\vec{v}_B$ being anti-parallel.