Physics - Rotation Question with Solution | TestHub

The ball has V', component of its velocity perpendicular to the length of rod immediately after the collision. u is velocity of COM of the rod and $\omega$ is angular velocity of the rod, just after collision. The ball strikes the rod with speed vcos53° in perpendicular direction and its component along the length of the rod after the collision is unchanged. Using for the point of collision. Velocity of separation = Velocity of approach

$$\Rightarrow \frac{3\mathrm{V}}{5}=\left(\frac{\omega \ell }{4}+\mathrm{u}\right)+{\mathrm{V}}^{\prime }$$

Conserving linear momentum (of rod + particle), in the direction ⟂ to the rod.

$$\mathrm{mV}\cdot \frac{3}{5}=\mathrm{mu}-{\mathrm{mV}}^{\prime }$$

Conserving angular moment about point ' D ' as shown in the figure

$$\begin{array}{rlr}& 0=0+\left[\mathrm{mu}\frac{\ell }{4}-\frac{\mathrm{m}{\ell }^2}{12}\omega \right]& \Rightarrow \mathrm{u}=\frac{\omega \ell }{3}\dots .(3)\end{array}$$

By solving

$$\mathrm{u}=\frac{24\mathrm{V}}{55},\mathrm{w}=\frac{72\mathrm{V}}{55\ell }$$

Time taken to rotate by $\pi$ angle $\mathrm{t}=\frac{\pi }{\omega }$

In the same time, distance travelled $={\mathrm{u}}_2.\mathrm{t}=\frac{\pi \ell }{3}$ Using angular impulse-angular momentum equation.

$$\begin{array}{rlr}& \int \mathrm{N}\cdot \mathrm{dt}\cdot \frac{\ell }{4}=\frac{\mathrm{m}{\ell }^2}{4}\cdot \frac{72\mathrm{V}}{55\ell }& \Rightarrow \int \mathrm{N}\cdot \mathrm{dt}=\frac{24\mathrm{mV}}{55}\}\text{ or }\{\begin{array}{l}\text{ using impulse }-\text{ momentum equation on Rod }\int \mathrm{Ndt}=\mathrm{mu}=\frac{24\mathrm{mv}}{55}\end{array}\end{array}$$