Question

A light, rigid rod of length l = 1 m joins two particles,with masses m1 =4 kg and m2=3 kg,at its ends. The combination rotates in the xy plane about the pivot through the center of the rod. Determine the angular momentum of the system
about the origin when the speed of each particle is 5 m/s.

Answer 1

First, find the angular velocity taking into consideration that both particles rotate 0.5 m from the axis of rotation:

$$\omega=\frac{v}{r}.$$

The moment of inertia of a single "sizeless" mass is

$$i=mr^2.$$

We have two particles, so, it is

$$I=i_1+i_2=m_1r^2+m_2r^2=r^2(m_1+m_2).$$

The angular momentum is

$$L=I\omega=(m_1+m_2)rv=17.5\text{ kg}\cdot\text{m}^2\text{/s}.$$