Recent content by frensel

Centripetal acceleration is ##a_c=\frac{v^2}{R}##. Since ##v=\frac{2\pi R}{T}##, where ##T## is the period, so ##a_c=\frac{4\pi^2 R}{T^2}##.

The initial position is ##x=0,y=0## at ##t=0##, the final position is ##x=X,y=h-Y## at ##t=T##.

I still didn't get it. It seems that it will not make any difference if I write down the equations in dynamical form. I assume ##x,y## are dynamical variables, ##X## and ##Y## are final horizontal and vertical displacement at final time ##t=T##. We have $$x=\frac{vYt}{R}$$...

I haven't said ##x## and ##y## are initial displacement... I said they are "horizontal and vertical displacement from the initial position to point A".

Sorry, I still can't figure out how to write down the proper equations... For the equation x=v\sin(a)t, it does not mean that x is time-dependent. x is the final horizontal displacement, therefore t is the the final time. With some final time t, the final horizontal and vertical displacement are...

x,y,a are not time-dependent. They are used to determined the initial velocity, therefore the projectile. For certain x, we can use R^2=x^2+y^2 and trigonometry to obtain y and a. There is only one solution of x (or y, a, and therefore initial velocity) so that the particle can arrive at point A.

I think it is correct. Noticed that the direction of the initial velocity is perpendicular to the radius of the circle.

Well...I can't understand your comment. x and h-y are the horizontal and vertical displacement from the initial position to point A. We want to know the final displacement of the particle, not the intermediate position. If x and y are determined, we get the initial velocity in horizontal...

In order to arrive at point A, the particle should be detached from the circular track at some point with fixed x and y and they are related to some angle a. While x and y are determined, we therefore can obtain the initial velocity of the particle and set the particle in parabola motion to point A.

Homework Statement As shown in the figure below, a particle is moving in a circle of radius R with constant speed v. At some location, the particle is detached from the circle and falls with a parabola path to point A. What is the horizontal range x of the projectile?Homework Equations Writing...

I got it, thx! \frac{\cos(\frac{x}{2})}{\cos{x}} = \frac{\sin(x)}{\sin(x)}\frac{\cos(\frac{x}{2})}{ \cos{x}} =\tan(x)\frac{\cos(\frac{x}{2})}{\sin(x)} using double-angle formula, we have \tan(x)\frac{\cos(\frac{x}{2})}{\sin(x)}...

Homework Statement How to convert \tan(x)\sin(\frac{x}{2})+\cos(\frac{x}{2}) to \frac{\tan(x)}{\sqrt{2(1-\cos(x))}} Homework Equations The Attempt at a Solution I can convert it to this form: \frac{\cos(\frac{x}{2})}{\cos(x)} \tan(x)\sin(\frac{x}{2})+\cos(\frac{x}{2})...

Hi, I follow the instruction in wikipedia to make a Cartesian diver (Coca Cola bottle + eyedropper), and it works well. I press the bottle, the eyedropper sinks; I release, it floats. However, if I turn the bottle upside down, the eyedropper sinks and it never floats again. That means the...

Thank you for your reply. I still have a question for you second point. If you heat strip A, A absorbs heat and some of the energy turn into the kinetic energy of A and some of it turn into the strain energy of A. Since strip A and strip B are in temperature difference now, heat flow from A to B...

Hi. If there are two objects A and B, the temperature of A is TA, the temperature of B is TB, and my questions are: 1). If TA = TB, their internal energy (which is the kinetic energy plus some other forms of energy) might not be the same. In this case, A and B are in thermal contact. Since TA...