Question

A uniform thin rod of mass $M$ and length $L$ is standing vertically along the $y-$axis on a smooth horizontal surface, with its lower end at the origin $(0,0)$. A slight disturbance at $t=0$ causes the lower end to slip on the smooth surface the positive $x-$axis, and the rod starts falling.
(i) What is the path followed by the centre of mass of the rod during its fall?
(ii) Find the equation of the trajectory of a point on the rod located at a distance $r$ from the lower end. What is the shape of the path of this point?

Answer 1

(i) Since, there is no horizontal force acting, center of mass of rod falls downwards.

(ii)

From the diagram, $x-$coordinate of Point P is:

$x=\frac{L}{2}\sin \theta -r\sin \theta$

y-coordinate of point P is:

$y=r\cos \theta$

Using ${\sin }^2\theta +{\cos }^2\theta =1$

$\frac{x^2}{(L/2-r{)}^2}+\frac{y^2}{r^2}=1$

The equation is that of an ellipse and hence, the shape of the path of this point is also an ellipse.