Question

A small disc of mass $m$ is sliding on a frictionless horizontal floor with a velocity $v_0$ in the positive y-direction of a coordinate system rigidly attached with the floor. At an instant $t = 0$, a horizontal force of constant magnitude $F$ begins to act on the disc in the positive x-direction. The force vector is made to rotate at a such constant rate that in time $\tau$ its direction becomes reversed, then the force is removed. Deduce a suitable expression for the velocity vector of the disc after the force is removed.

Answer 1

$\vec{v}_f = \left( v_0 + \frac{2F\tau}{m\pi} \right) \hat{j}$

Answer 2

The force vector rotates uniformly from positive x-direction to negative x-direction in time $\tau$. This implies its angular velocity is $\pi/\tau$. The force vector is expressed as $\vec{F}(t) = F \cos(\frac{\pi t}{\tau}) \hat{i} + F \sin(\frac{\pi t}{\tau}) \hat{j}$. The impulse imparted by this force is $\int_0^\tau \vec{F}(t) dt$. Integrating the x-component of the force from $0$ to $\tau$ yields $0$, while integrating the y-component yields $\frac{2F\tau}{\pi}$. Thus, the total impulse is $\frac{2F\tau}{\pi} \hat{j}$. By the impulse-momentum theorem, $m\Delta\vec{v} = \text{Impulse}$, so $\Delta\vec{v} = \frac{2F\tau}{m\pi} \hat{j}$. Adding this change in velocity to the initial velocity $\vec{v}_0 = v_0 \hat{j}$ gives the final velocity $\vec{v}_f = \left( v_0 + \frac{2F\tau}{m\pi} \right) \hat{j}$.