Question

A uniform rod of length 20cm is freely pivoted to the center. A gum of mass 0.2kg moving at a speed of 10ms^-1 strikes and sticks to one end of the rod. The rod rotates horizontally. The moment of inertia of the rod and the gum about the pivot is 0.15kgm^2
Calculate angular momentum of gum?
Calculate final angular velocity?
If the length of rod is doubled and moment of inertia is 0.20,calculate the new angular velocity of system.

Answer 1

The angular momentum of gum:

$$L=mv(0.5l)=0.5(0.2)(10)(0.2)=0.2\frac{kgm^2}{s}$$

From the conservation of angular momentum:

$$I\omega=L$$

$$\omega=\frac{L}{I}=\frac{0.2}{0.15}=1.3\frac{rad}{s}$$

If the length of rod is doubled and moment of inertia is 0.20:

$$\omega'=\frac{2L}{I}=2\frac{0.2}{0.2}=2\frac{rad}{s}$$