Question

A rod of mass M = 146 g and length L = 47 cm can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m = 19 g, moving with speed V = 5 m/s, strikes the rod at angle θ = 29° a distance D = L/2 from the end and sticks to the rod after the collision.What is the angular speed ωf of the system immediately after the collision, in terms of system parameters and I?

Answer 1

Answer:

$\omega = \frac{mv\frac{L}{2}\sin(29^\circ)}{\frac{1}{3}ML^2 + m(\frac{L}{2})^2} = 0.91~{\rm rad/s}$

Explanation:

The angular speed of the system can be found by conservation of angular momentum. Since the ball and the rod are stick together, the collision is completely inelastic, ergo kinetic energy is not conserved.

$L_{initial} = L_{final}\\m\vec{v}\times \vec{r} = I\omega$

The moment of inertia of the combined objects is equal to the sum of moment of inertia of the separate objects.

$I = \frac{1}{3}ML^2 + m(\frac{L}{2})^2$

The cross product in the left-hand side can be written as a sine of the angle.

Therefore;

$mv\frac{L}{2}\sin(29^\circ) = (\frac{1}{3}ML^2 + m(\frac{L}{2})^2)\omega\\(19\times 10^{-3})(5)(\frac{47\times 10^{-2}}{2})\sin(29^\circ) = (\frac{1}{3}(146\times 10^{-3})(47\times 10^{-2})^2 + (19\times 10^{-3})(\frac{47\times 10^{-2}}{2})^2)\omega\\\omega = 0.91~{\rm rad/s}$In terms of system parameters:

$\omega = \frac{mv\frac{L}{2}\sin(29^\circ)}{\frac{1}{3}ML^2 + m(\frac{L}{2})^2}$