Question

Figure shows a rod of length / resting on a wall and the floor. Its lower and A pulled towards left with a constant velocity u. As a result of this, end A starts moving down along the wall. Find the velocity of the other end B downward when the rod makes an angle $\theta$ with the horizontal.

Answer 1

Let us first find the relation between the two displacements. then differentiate with respect to time. Here if the distance from the corner to the point A is x and that up to B is y, the ;eft velocity of point A can be given as $v_A=dx/dt$ and that of B can be given as $v_A=-dy/dt$ (negative sign indicates y decreasing).

If we relate x and y: $x^2+y^2=l^2$

Differentiating with respect to t, $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$

$\Rightarrow$ ${xv}_A={yv}_B\Rightarrow xu={yv}_B\Rightarrow v_B=u\frac{x}{y}u\cot \theta$

Alternatively :

In cases where distance between two points is always fixed, we can say the relative velocity of one point of an object with respect to any other point of the same object in the direction of the line joining them will always remain zero, as their separation always remains constant.

Here, in the above example, the distance between the points A and B of the rod always remains constant; thus, the two points must have the same velocity components in the direction of their line joining, i.e., along the length of the rod.

If pint B is moving down with velocity $v_B$ its component along the length of the rod is $v_b$, its component along the length of the rod is $v_B\sin \theta$. Similarly, the velocity component of point A along the length of rod is $u\cos \theta$. Thus, we have $v_B\sin \theta =u\cos \theta$ or $v_B=u\cot \theta$.