Recent content by Father_Ing

The hint says the following: "Since the cords are inextensible, every particle of a cord must be in circular motion about the point where it is affixed to the ceiling. Therefore, the velocities of the points where the cords are leaving the disc are perpendicular to the string" Due to the fact...

If we look at the 2nd mirror, could we see the virtual object?

I'm currently confused in determining whether an image formed by the 1st mirror (the left one) is a real or virtual object for the 2nd mirror. Here is the solution manual: This is what I have in my mind: Since the object is located between the focus and radius point of the first mirror, the...

But, all three of them depends on time (because if it is not, I don't think we can differentiate it with respect to time).

$$L = \frac {mv^2}{2} - mgy$$ It is clear that ##\dot{x}=\dot{\theta}L## and ##y=-Lcos \theta##. After substituting these two equations to Lagrange equation, we will get the answer by simply using this equation: $$\frac {d} {dt} \frac {∂L}{∂\dot{\theta}} - \frac {∂L}{∂\theta }= 0$$ But, What if...

Why the other end(the one that is standing still) does not accelerate? How can we know that the tension in this end is zero?

"Let x be the distance your hand has moved.Then, x/2 is the length of the moving part of the chain, because the chain gets "doubled up""

You mean the force that the hand gives? But, that force is not applied to the end that is standing still.

Ah.. I should have said the hand moved at L distance in the example. My bad. (I already edited it) Isn't there any tension between each part? If the rest of the chain (such as the other end of the chain) does not move at all, what force that cancel the leftward tension on this part?

In the book, it is stated that if your hand move a distance x, then x/2 is the length of the moving part of the chain because the chain gets “doubled up.” as in the image below. I don't get the meaning of this. For example, if our hand move L metres from initial position, shouldn't the moving...

Consider a rocket with mass ##m## in space is going to move forward. In order to do so, it needs to eject mass backwards. Let the mass that is ejected has velocity ##u## relative to the rocket. What is the equation for the final velocity? It is said that after ##dt## second, the rocket will...

In the solution manual, it says that: the resultant of friction force is ##<= kmg##, hence $$m\sqrt{\omega_t^2 + (\frac {v^2} {R})^2} <= kmg$$ and from this equation, we will get $$v^2 <= R \sqrt{(kg)^2 -\omega_t^2}$$ which will make ##v_{max}^2= R \sqrt{(kg)^2 -\omega_t^2}## Finally, they...

Alright! Thanks.

Yep! I have made it through the 2nd question. I still have some problems regarding the concepts, though. So, each particle has a force ##dN## exerted on it, and the value of ##dN## differs from each particle?

I am a bit confused.. If we see the the rod as a rigid body that is formed by many particles, N is exerted only to the particle at the end of the rod.In my opinion, N only affects the acceleration of the particle this point (the pivot, or we can also say, the end of the rod). However,in the...