Question

Figure shows a rod of length l resting on a wall and the floor. Its lower end A is pulled towards left with a constant velocity u. As a result of this, end B 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.

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

$x^2+y^2=l^2$

Differentiating with respect to t,

$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\Rightarrow x{v}_A=y{v}_B\Rightarrow xu=y{v}_B\Rightarrow v_B=u\frac{x}{y}=ucot\theta$

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 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 point B is moving down with velocity $v_B$, 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 $\theta$.