Calculating Velocity at a Specific Point on a Curved Slit

AI Thread Summary
The discussion centers on calculating the velocity at a specific point on a curved arch defined by a sine function. The initial attempt to find the vertical velocity component (vy) using the derivative of the position equation led to confusion regarding the correct application of the chain rule. The participant initially miscalculated the resultant velocity and mistakenly thought they were asked for acceleration instead of velocity. Clarifications were provided about the nature of the cosine function and its domain, alleviating concerns about the argument exceeding one. The conversation highlights the importance of correctly applying calculus principles in physics problems.
Melawrghk
Messages
140
Reaction score
0

Homework Statement


The shape of the arch is defined by a sine function with a = 0.2 and the width of the arch is 1m. The velocity v is constant at 2m/s. What is the velocity at point P when x=0.25m?

arc.jpg


The Attempt at a Solution


So since v is just in the x-direction, I figured that's the vx speed.
Then I came to having to find an equation to vy and this is where I get stuck. I remember from high school physics that equation of velocity can be essentially found by taking a derivative of the position equation.

In this case I get:
vy=y'=0.2*pi*cos(pi*x)

From which I can find that at x=0.25m, vy = 0.444288 m/s
But that combined with the horizontal speed of 2m/s gives me a resultant velocity of 2.05ish, instead of 2.19 that I'm supposed to get. Help please?
 
Last edited:
Physics news on Phys.org
It is asking for acceleration, not velocity. You are adding apples to oranges.

BTW, is that the exact wording of the problem? - it is appauling.
 
Yeah, that's the exact wording.

Sorry, the question is a two parter. I'll fix that. I just decided to not even tackle acceleration if I can't figure out velocity.
 
Where did I see acceleration?! I must be going mad. I am so sorry. Anyway, your velocity calculation is almost correct. Your problem is that vy is the derivative of y wrt t, not x. Think chain rule.
 
Oh thank you, thank you! :) That allows me to get acceleration as well. Yay!
 
Except... Now I'm doing a question that's basically the same, except

y=0.23sin(pi*x) and Vx=2.4m/s and we have to find velocity at x=0.32

I did the same thing as before and got:
Vy = 0.552*pi*cos(2.4*pi*t)

But the thing in the brackets ends up being >1 and that can't be. What did I do? :S
 
Melawrghk said:
Vy = 0.552*pi*cos(2.4*pi*t)

But the thing in the brackets ends up being >1 and that can't be. What did I do? :S
What thing in brackets? Do you mean 2.4πt? What's wrong with that being >1?
 
Yes that, but cos can't be >1...
 
The cosine is not >1; its argument is >1. The domain of cos(x) is x any real number, (and actually this can be generalized even further). I think you're worried about the range. Don't worry about that; your calculator will take care of it.
 
Back
Top