# What is the distance traveled

the red dots represent a side view of path traveled, F is downward force and the tool used here is a pen placing parallel to the coin

I have newly started to study mechanical physics. based on study, I conduct a simple experiment. But unfortunately I am unable apply the laws in reality.

Experiment: I placed a coin (2 Indian Rupee coin) with radius $r$ positioned flat part of coin parallel to base of my laptop. Also placed at edge of base such that $x$ (mm) of diameter of coin is supported by laptop while $2 r - x$ is free, unsupported with the coin in a balanced position. Now at $x+y\, <\, 2 r$" measured along diameter of coin toward the free end I applied a random force perpendicular to laptop base.

Now my question is , how can we compute(formulate) distance $z$ traveled by coin measured from center of coin at start point to center of coin at the place it stopped in terms of known variables mentioned below.

My attempt: To me the known values are: $r$,$x$,$y$,density of coin($\rho$),width(or height) of coin($w$), time it took to stop($t$).

Here for simplicity I have not considered the torque generated, air resistance, and visualized the traveled path as a simple parabolic path. Now I can measure the distance traveled $z = v_0t+\frac{1}{2}gt^{2}$. where g is acceleration due to gravity.

But how can I find the initial velocity $v_0$. I know $v_0$ is not zero, since the situation here is not equivalent to a freely falling scenario.

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This is a very complex situation that will be quite difficult to analyze for someone who is new to the subject. The vertical movement comes from transfering the downward force of your finger to the center of mass of the coin via a lever with the fulcrum at the corner of the laptop. Rather than calculating where the coin will land, you might try measuring the time of flight and calculating the downward force imparted by your finger. – AdamRedwine May 22 '13 at 18:19
@AdamRedwine: Can you please elaborate the complexity of situation. Mean while I will try what you have suggested. – smslce May 22 '13 at 20:21
A sketch would be greatly appreciated. I have trouble understanding the situation. – ja72 May 22 '13 at 21:12
Where exactly is the pen hitting the coin in relation to the center of the coin? – ja72 May 23 '13 at 4:32
@ja72 : It's at distance $y$ measured from edge of laptop to the force applied by pen. So it's actually at a distance of $(x-r)+y$ from the center of coin, also the force applied is not at a single point on that distance. Since it's a pen held parallel to coin and also parallel to the laptop edge, the force is applied from one end of coin to the other end. – smslce May 23 '13 at 4:45

1. Impact: The pen applies an impulse over a short period of time at $b$ and the coin pivots about $a$ as the center of gravity starts to lift off. Initial speed is given from the impulse magnitude and the geometry.