a ladder AP of length 5m inclined to a vertical wall is slipping over a horizontal surface with velocity 2m/s , when A is at a distance 3m from ground . what is the velocity of centre of mass at this moment?

Question

a ladder AP of length 5m inclined to a vertical wall is slipping over a horizontal surface with velocity 2m/s , when A is at a distance 3m from ground . what is the velocity of centre of mass at this moment?

Answer 1

To find the velocity of the center of mass of the ladder, we can consider the ladder as a system of two particles - one at point A and the other at the center of mass C. We can track the motion of point A and use the concept of relative motion to find the velocity of the center of mass.

Let's first consider the motion of point A. We are given that the ladder is slipping over a horizontal surface with a velocity of 2 m/s. This implies that the velocity of point A (VA) is also 2 m/s and is parallel to the surface.

Next, we can calculate the velocity of point A in terms of its components relative to the center of mass C.

Let's assume that the ladder makes an angle θ with the horizontal surface. We know the distance of point A from the center of mass (AC) is half the length of the ladder, which is 2.5 m.

The velocity of point A relative to the center of mass (VA') can be given by:

VA' = VA - VC

Since the components of VA and VC in the horizontal direction are equal, we can write:

VA' = 2 - VC

Now, let's find VC. Since the ladder is inclined to the vertical wall, we can use trigonometry to determine VC.

The ladder forms a right triangle with the vertical wall, where AC is the hypotenuse. Using the given information, we can apply the Pythagorean theorem:

AC^2 = 3^2 + 2.5^2
AC^2 = 9 + 6.25
AC = √15.25
AC ≈ 3.91 m

Now, we can find θ using trigonometry:

sin(θ) = opposite / hypotenuse
sin(θ) = 2.5 / 3.91
θ ≈ sin^(-1)(2.5/3.91)
θ ≈ 38.4 degrees

Since the ladder is slipping, θ is changing with time. But at the specific moment when A is at a distance of 3 m from the ground, we can use this value.

Now, let's find VC using the angle θ:

VC = VA * sin(θ)
VC = 2 * sin(38.4 degrees)
VC ≈ 1.23 m/s

Finally, we can find VA' by substituting the values:

VA' = 2 - VC
VA' = 2 - 1.23
VA' ≈ 0.77 m/s

Therefore, the velocity of the center of mass at this moment is approximately 0.77 m/s.

a ladder AP of length 5m inclined to a vertical wall is slipping over a horizontal surface with velocity 2m/s , when A is at a distance 3m from ground . what is the velocity of centre of mass at this moment?

i-3/4j

To find the velocity of the center of mass of the ladder, we can consider the ladder as a system of two particles - one at point A and the other at the center of mass C. We can track the motion of point A and use the concept of relative motion to find the velocity of the center of mass.