Answer to Question #111634 in Classical Mechanics for Isac Colta

Question

A wooden rod, 40 cm long, with uniformly distributed mass of 1 kg lies on a flat horizontal
frictionless surface. A force, as shown in the figure, acts on the rod 10 cm from its center.
Assume that the force acted perpendicular to the rod throughout the duration of the contact.
Also assume that the axis of rotation is at the center of the rod, perpendicular to the horizontal
surface. As a result of the impact, the rod will start rotating and translating at the same time
(analogous to the ‘rolling motion’). Find the linear and angular velocities of a point P located at
the bottom end of the rod.

Answer 1

Initially the rod did not have angular and linear velocity. According to Newton's second law, during the impact

$F=m\frac{dv}{dt},\\ mdv=Fdt.$

Integration gives final velocity of the center of mass, while the Fdt term can be found from the graph attached with the condition of the problem:

$mv_f=1000\cdot0.002,\\ v_f=2\text{ m/s}.$

Since the rod will rotate freely, it will rotate around its center of mass. The moment of inertia:

$I=\frac{mL^2}{12}.$

The torque is force times lever. Force is 1 kN, lever is 10 cm:

$\tau=Fl.$

This torque creates angular acceleration:

$\tau=I\alpha, \\ Fl=\frac{mL^2}{12}\frac{d\omega}{dt},\\ mL^2d\omega=12Fldt,\\ mL^2\int_0^{\omega_f} d\omega=12Fl\int_0^t dt,\\ \omega_f=\frac{6Flt}{mL^2}=\frac{6\cdot10^3\cdot0.1\cdot0.004}{1\cdot0.4^2}=15\text{ rad/s}.$

As the rod rotates, the linear velocity of the point P only is

$v=\omega_f\frac{L}{2}=3\text{ m/s},$

the stationary observer standing on that frictionless surface will see that the point P moves at

$v_p=v_f+v\text{ cos}(\omega_ft)=2+3\text{ cos}(15t).$