How much time do I have to catch a coin?

In summary, there is a confusion about velocities in this problem and whether the coin's initial velocity is affected by the movement of the walkway. From Philipp's perspective, the coin's initial velocity is zero and its transfer kinetic energy is also zero. However, from the speaker's perspective, the coin's initial velocity is not zero due to the movement of the walkway. This leads to a discrepancy in the equations, but it is resolved when considering the total distance the coin must travel.
  • #1
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Homework Statement
When riding up the inclined moving walkway of inclination ##α## and length ##l## a coin drops out of Philipp’s pocket when he is exactly in the middle of it. It falls into one of the grooves on the walkway and starts rolling down without slipping. How much time does Philipp have to catch the coin before it falls under the bottom edge of the walkway? The velocity of the moving walkway is ##v##.
Relevant Equations
I would use an euqation for a rotational kinetic energy ##\frac 12 I{\omega}^2## and an equation for a transfer kinetic energy ##\frac 12 mv^2##. ##I=\frac 12 mR^2##.
I am a bit confused with velocities in this problem. From Philipp's view, the coin's initial velocity is zero, so its transfer kinetic energy is also zero. When I am standing on a non-moving ground, is the coin's initial velocity ##v## in direction the walkway is moving? But won't I get then different times?

From Philipp's view: ##\frac 12 l=\frac 12 at^2##
From my view: ##\frac12 l=\frac 12 at^2-vt##

Where do I do a mistake?
 
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  • #2
There are four quantities there. You've assumed correctly that ##a## and ##t## are the same in both frames. And, that ##v## is different in the two frames (##v = 0## in Philipp's frame). What about ##l##? Is that the same in both frames?
 
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  • #3
PeroK said:
There are four quantities there. You've assumed correctly that ##a## and ##t## are the same in both frames. And, that ##v## is different in the two frames (##v = 0## in Philipp's frame). What about ##l##? Is that the same in both frames?
Now I understand, in Philipp's frame, the total way the coin must travel is ##\frac 12 l+vt##, so then the equations are equivalent.
 
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FAQ: How much time do I have to catch a coin?

How much time do I have to catch a coin dropped from a certain height?

The time you have to catch a coin depends on the height from which it is dropped. You can use the equation of motion under gravity: \( t = \sqrt{\frac{2h}{g}} \), where \( h \) is the height and \( g \) is the acceleration due to gravity (approximately 9.8 m/s²).

What factors affect the time I have to catch a falling coin?

The primary factors are the height from which the coin is dropped and the acceleration due to gravity. Air resistance is generally negligible for a small object like a coin over short distances.

Can I increase the time I have to catch a coin?

Yes, you can increase the time by either increasing the height from which the coin is dropped or by slowing the coin's descent, possibly by using a medium with higher air resistance, although this is less practical in everyday scenarios.

How does the mass of the coin affect the time to catch it?

The mass of the coin does not affect the time it takes to fall, as all objects accelerate at the same rate under gravity, assuming air resistance is negligible.

Is there a way to calculate the exact time to catch a coin without using a stopwatch?

Yes, you can calculate the time using the formula \( t = \sqrt{\frac{2h}{g}} \). Measure the height from which the coin is dropped and use 9.8 m/s² for \( g \) to find the time \( t \) in seconds.

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