Question

a rod leans against a vertical wall as shown in figure. the other end of the rod is on the horizontal floor. point A is pushed downwards with constant velocity path of the centre of the rod is?

Answer 1

The centre of the rod will always be at position \[\left(\frac{A}{2},\frac{B}{2}\right)\]Since the rod, wall and floor always make a right-angled triangle, and the hypotenuse of the triangle is constant (length of the rod doesn't change), you can say that\[A^2 + B^2 = H^2\], you can then divide everything by 4 (2 inside the square):\[\left(\frac{A}{2}\right)^2 + \left(\frac{B}{2}\right)^2 =\left(\frac{H}{2}\right)^2\]Since A and B are variables, and H is a constant, what you have looks eerily similar to the equation for a circle!

Answer 2

no the answer is wrong
her is the correct answer

let l be the lenght of the rod.
the angle of rod with x-axis at some instant of time. co-ordinates of centre of rod at this instant are

x=l/2 cos(theta)

y = l/2 sin(theta)

squaring and adding the equations u get

x^2 + y^2 = (L^2)/4

this is an equation of a circle with radius l/2 and centre at origin

Answer 3

That's the same as the answer I gave :)

Answer 4

u have given that that the centre is constant
but i tell the centre follows a path of a circle

Answer 5

I've said the centre follows the path of the circle also: \[\left(\frac{A}{2}\right)^2 + \left(\frac{B}{2}\right)^2 =\left(\frac{H}{2}\right)^2\] is the equation of a circle.

Answer 6

I'm taking A and B as being variables on the X and Y axes, not fixed points.

Answer 7

ok ok i give up u might be correct

Answer 8

Well I had \[\left(\frac{A}{2}\right)^2 + \left(\frac{B}{2}\right)^2 =\left(\frac{H}{2}\right)^2\]You had \[\left(x\right)^2 + \left(y\right)^2 =\left(\frac{L}{2}\right)^2\]The only difference is the name of the variables :P

Answer 9

yeah its just a confusion