Question

Question for ❤️PHYSICS NEWTONS ❤️

EXPLAIN IN DETAILS...!!!​

Answer 1

Answer:

answer is option 1. Check the solution

Answer 2

Answer:

Given:

Velocity of slipping of uppermost part of ladder is v1 , while the lowermost part is slipping with v2.

To find:

Velocity of centre of mass of rod at that instant.

Concept:

We shall calculate the velocity of centre of mass of the rod with respect to X and Y axis . After that , we can perform vector addition to get the resultant velocity.

Calculation:

Let's consider the mass of whole rod be located at the 2 ends (i.e considering the whole rod to be 2 point masses - each point mass be m)

X component velocity at centre of mass:

$\displaystyle \: v_{x} = \dfrac{ \Sigma(mv)}{ \Sigma(m)}$

$\displaystyle \: v_{x} = \dfrac{ \{m(v1) +( m \times 0) \}}{ (m + m)}$

$\displaystyle \: v_{x} = \dfrac{(v1)}{2}$

Similarly , The Y component of velocity at centre of mass will be

$\displaystyle \: v_{y} = \dfrac{(v2)}{2}$

As per vector addition , we can say that :

$\: v_{net} = \sqrt{ {v_{x}}^{2} + {v_{y}}^{2} }$

$= > \: v_{net} = \sqrt{ { \bigg \{\dfrac{ (v1)}{2} \bigg \} }^{2} + { \bigg \{ \dfrac{(v2)}{2} \bigg \} }^{2} }$

$= > \: v_{net} = \frac{1}{2} \sqrt{ {(v1)}^{2} + {(v2)}^{2} }$

So final answer :

$\boxed{ \huge{ \red{ \sf{ \: v_{net} = \frac{1}{2} \sqrt{ {(v1)}^{2} + {(v2)}^{2} } }}}}$