Mechanics question derivations terminal velocity (quadratic case)

[Genericcoder]

hi so basically my question is derivation of derivation of the quadratic case of the terminal velocity


I have attached the pictures in picture 1 I don't know I think the book has error in his derivation @ equation 2.4.13 shouldn't it be + instead of negative ??

& in second picture their is something I don't understand two stuff acctually first is in equation 2.4.15
how did he come up with 1/2 * dv^2 / dy? and my second question is
du/dy = -1/vt^2 * dv^2/dy how did he come up with dv^2/dy ??

thanks alot.

 

[maajdl]

Yes the sign is wrong.
And dv/dt = dv/dy dy/dt = dv/dy v
since v = dy/dt .
And since dv²/dt = 2 v dv/dt, the last result follows.

 

[Genericcoder]

could u explain this line a little more ->dv²/dt = 2 v dv/dt, the last result follows?

 

[maajdl]

read:

dv²/dy = 2 v dv/dy

which implies

v dv/dy = 1/2 dv²/dy

 

[HalfLostAndSearching]

I am always happy to assist with questions related to mechanics and derivations. In order to properly address your inquiries, I will need to see the pictures you have referenced. However, I can provide a general explanation of the quadratic case of terminal velocity.

In the case of an object falling through a fluid, the terminal velocity is the maximum velocity that the object will reach due to the balance of two forces: gravity pulling the object down and air resistance pushing the object up. The quadratic case refers to the situation where the air resistance is proportional to the square of the object's velocity.

To derive the equation for terminal velocity in the quadratic case, we can start with the equations of motion for the object:

F = ma = mg - bv^2

Where F is the net force, m is the mass of the object, g is the acceleration due to gravity, b is the drag coefficient, and v is the velocity of the object. At terminal velocity, the net force is equal to zero, so we can set the equation equal to zero and solve for v:

0 = mg - bv^2

v = √(mg/b)

This is the equation for terminal velocity in the quadratic case. Now, let's address your specific questions about the derivations in the pictures you have referenced.

In picture 1, it appears that you are questioning the sign in equation 2.4.13. Without seeing the picture, I cannot say for certain, but it is possible that the negative sign is correct. In some cases, the direction of the force or velocity may need to be reversed in order to account for the direction of the vector. I would need to see the full derivation in order to confirm.

In picture 2, you are asking about the derivation of 1/2 * dv^2 / dy and dv^2 / dy. These terms come from the chain rule in calculus. In this case, the velocity (v) is a function of the position (y), so when we take the derivative of v with respect to y, we use the chain rule to get 1/2 * dv^2 / dy. Similarly, when we take the derivative of v^2 with respect to y, we use the chain rule again to get dv^2/dy. I hope this explanation helps to clarify your questions.

In summary, the quadratic case of terminal velocity is derived by balancing the forces acting on an object falling through a