Question

Consider a situation in which a uniform gravitation field is present below line AB and
no gravity above line AB. Then the speed with which a rod shown in the figure will
strikes earth surface when released from the position shown is:
-В
SZ
earth surface
Oſgl
O2/gl
03/gl
4/gl​

Answer 1

Answer:

Consider a situation in which a uniform gravitation field is present below line AB and

no gravity above line AB. Then the speed with which a rod shown in the figure will

strikes earth surface when released from the position shown is:

-В

SZ

earth surface

Oſgl

O2/gl

03/gl

4/gl

Answer 2

The speed at which the rod strikes the earth surface when released from the position is  3$\sqrt{gl}$

Given:

Gravitation field is present below line AB, no gravity above line AB.

To Find:

The speed with which a rod shown in the figure will strikes earth surface when released from the position

Solution:

To get the value of the initial position when it strikes the earth surface

is given by,

a = $\frac{\frac{m}{l}yg }{m}$

= $\frac{gy}{l}$

The limit tends from 0 to u,

We get,

∫ 0 to u (vdv) = $\frac{g}{l}$ ∫ 0 to u (ydy)

u = $\sqrt{gl}$.......(1)

$v^{2} = u^{2} +2g(4l)$

u = 3$\sqrt{gl}$

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