Maximum Acceleration: I Got the Correct Answer

[rashida564]

Homework Statement

Position of an object is given by y(X)=sin(x), and moves with constant speed v. Find the maximum speed given that, the acceleration at the top shouldn't exceed g.

Relevant Equations

a=dv/dt

I got the correct answer which is √g. I solved it as if it were a circular motion with radius of 1. But got no idea why my solution did work.

 

[berkeman]

Homework Statement:: Position of an object is given by y(X)=sin(x), and moves with constant speed v.

Please show your detailed work so we can check it.

Also, this problem statement is confusing to me. Is it saying that v(t) is constant, and the position y(x) follows a sinusoidal path? How can you find the "maximum speed" if the speed is constant?

 

[haruspex]

Is it saying that v(t) is constant, and the position y(x) follows a sinusoidal path? How can you find the "maximum speed" if the speed is constant?

It says the speed is a constant v. Whether you consider it v(t) or v(x) is immaterial. It asks for the maximum value of a constant given a constraint, not the maximum over time.

 

[berkeman]

Thanks @haruspex -- I still am not understanding, but will just watch to see how the OP responds. Guess I'll learn something.

 

[haruspex]

Homework Statement:: Position of an object is given by y(X)=sin(x), and moves with constant speed v. Find the maximum speed given that, the acceleration at the top shouldn't exceed g.
Homework Equations:: a=dv/dt

I got the correct answer which is √g. I solved it as if it were a circular motion with radius of 1. But got no idea why my solution did work.

I think you just got lucky. What if it had been A sin(ωx)?

 

[haruspex]

Thanks @haruspex -- I still am not understanding, but will just watch to see how the OP responds. Guess I'll learn something.

It's like asking for the maximum safe cornering speed on a non-uniformly curved track.

 

[rashida564]

I think you just got lucky. What if it had been A sin(ωx)?

can you give me a hit of what should I do

 

[haruspex]

can you give me a hit of what should I do

What determines the acceleration at a given point on a curve if the speed is constant?

 

[rashida564]

The direction of the acceleration. It must be tangential to the curve

 

[rashida564]

Is it true this way because it will be a rotating vector, or I just got it by pure luck again
not sure though why should it v times v

 

[etotheipi]

You know that $y = \sin{x}$ describes the trajectory of the particle, and you want to find something relating to the acceleration in the $y$ direction, $\frac{d^{2}y}{dt^{2}}$. What happens if you try finding the second derivative of $y$ with respect to $t$? Then, what values can you substitute into this expression given the critical condition in the question?

 

[haruspex]

The direction of the acceleration. It must be tangential to the curve

No, no. Tangential acceleration would change the speed.

 

[haruspex]

something relating to the acceleration in the y direction

Not in general, but in this case yes. E.g. consider rotating the axes a little.

 

[haruspex]

Is it true this way because it will be a rotating vector, or I just got it by pure luck again
not sure though why should it v times v

You'd need to justify dv/dt = v². Where did that come from? Remember |v| is given as constant.

 

[rashida564]

that's from the rotation of axis. Like any vector. dA/dt= A*w. where w is the rate in which it changes it's direction. Not sure why this vector change it's direction by magnitude of v

 

[haruspex]

Like any vector. dA/dt= A*w. where w is the rate in which it changes it's direction.

If you want to justify it in terms of vectors you'll need to write it appropriately to be understood.
Do you mean that if $|\vec A|$ is constant then $|\frac{d\vec A}{dt}|=|\vec A|\omega$ where ω is .. what algebraically ?

 

[ehild]

Homework Statement:: Position of an object is given by y(X)=sin(x),

I got the correct answer which is √g. I solved it as if it were a circular motion with radius of 1. But got no idea why my solution did work.

It is not a circular motion.
Imagine it is a car going up a hill with constant speed v. The horizontal displacement of the car is x, the vertical one is y. The shape of the hill is y=sin(x). Determine the x and y components of acceleration at the top of the hill.

 

[rashida564]

If you want to justify it in terms of vectors you'll need to write it appropriately to be understood.
Do you mean that if $|\vec A|$ is constant then $|\frac{d\vec A}{dt}|=|\vec A|\omega$ where ω is .. what algebraically ?

Yes sir sorry for my messy writing, this what I am trying to figure out, the left hand side is " |d→Adt||dA→dt| clearly g, the $|\vec A|\$ is v, but I don't get what's w

 

[haruspex]

Yes sir sorry for my messy writing

So how do you get $|\frac{\vec{dv}}{dt}|=|\vec v|^2$?

 

[rashida564]

w must be equal to v, but I don't know how to define w. like what's it's formula

 

[haruspex]

w must be equal to v, but I don't know how to define w. like what's it's formula

What do you know about centripetal acceleration?

 

[rashida564]

It's to the center and equal to v^2/r

 

[haruspex]

It's to the center and equal to v^2/r

And what is its direction relative to the velocity?