What is the Path of a Point Mass with Perpendicular Acceleration?

[nand_1]

Hi,

If i had a point mass moving with a velocity V in a certain direction (with constant speed), and a constant acceleration (or force) was acting perpendicular to it's velocity vector. From my understanding, the point mass' speed will not change, however it will begin to take a circular path?

Now how would I trace the path the point takes if the acceleration was no longer constant but was time varying?

For example, if the acceleration acting perpendicular to velocity vector was a*sin(wt). Is there a simple way of showing the path the point mass would take if it's speed remains constant?

Regards,

 

[nand_1]

^Bump^

 

[rcgldr]

I doubt there's a simple way. You'd need to integrate (twice) to determine position versus time (which would be the path).

 

[SAINATHAN]

when a perpendicular force acting how it can take circular path.

 

[Studiot]

The replies are probably a bit vague because your query is ill defined.

What do you mean by a constant velocity v?
Do you mean like a ship moving with the tide or do you mean like an object moving freely in space in accordance with Newtons first law?

As soon as the acceleration is applied the object's velocity will have another component.
Thus it will have a resultant velocity, different from v, say v'

Is the acceleration now to be perpendicular to v or v'?

 

[sylas]

For example, if the acceleration acting perpendicular to velocity vector was a*sin(wt). Is there a simple way of showing the path the point mass would take if it's speed remains constant?

If the acceleration is always perpendicular the velocity vector, then the speed WILL remain constant, no matter how the acceleration varies.

You can calculate the path as long as you are limited to movement in two dimensions (movement in a plane) and have a fixed convention for whether +ve acceleration turns to the right or to the left. You don't need to assume constant speed; that will fall out automatically from the calculation.

Cheers -- sylas

 

[nand_1]

If the acceleration is always perpendicular the velocity vector, then the speed WILL remain constant, no matter how the acceleration varies.

You can calculate the path as long as you are limited to movement in two dimensions (movement in a plane) and have a fixed convention for whether +ve acceleration turns to the right or to the left. You don't need to assume constant speed; that will fall out automatically from the calculation.

Cheers -- sylas

Sorry about that I worded it wrong but that definitely makes sense.

Basically I want to calculate the path in a 2D plane, however, there doesn't seem to be an easy way of doing this for time varying perpendicular acceleration. I have plotted the path taken in a MATLAB but i was just wondering if i could have done it manually first.

Studiot: The acceleration is always perpendicular to the current velocity vector and the object is moving freely in 2D space.

Thanks for all the replys.

 

[sylas]

Basically I want to calculate the path in a 2D plane, however, there doesn't seem to be an easy way of doing this for time varying perpendicular acceleration. I have plotted the path taken in a MATLAB but i was just wondering if i could have done it manually first.

You can set up differential equations as follows. Let x and y be the location of the particle at time t, in co-ordinates chosen so that the boundary conditions at t=0 are

<br /> \begin{align*}<br /> x &amp; = 0 \\<br /> y &amp; = 0 \\<br /> \frac{dx}{dt} &amp; = 1 \\<br /> \frac{dy}{dt} &amp; = 0<br /> \end{align*}​

Let f(t) be a function that gives the magnitude of acceleration at time t. You also have that acceleration is perpendicular to velocity. This gives two differential equations.

<br /> \begin{align*}<br /> (\frac{d^2x}{dt^2})^2 + (\frac{d^2y}{dt^2})^2 &amp; = f(t)^2 \\<br /> \frac{dx}{dt} \frac{d^2y}{dt^2} + \frac{dy}{dt}\frac{d^2x}{dt^2} &amp; = 0<br /> \end{align*}​

Your problem reduces to finding a solution for this set of differential equations.

Cheers -- sylas

 

[nand_1]

Thanks for that Sylas.