# Curillinear Motion

• Sep 28th 2010, 09:05 PM
Techro
Curillinear Motion
I am attempted this question a few different ways and I am stuck

PROBLEM:
An object follows a circular path with a radius of 30meters. The acceleration of the object along the path is given by the equation a = 6s where a is in m/s^s and s is the distance traveled along the path in meters.
If the initial velocity is 9m/s, find the magnitude of the acceleration when it has traveled 24meters.

I thought I was doing it right.

ATTEMPT:
• Sep 29th 2010, 05:07 AM
Ackbeet
I have a couple of comments.

1. The acceleration a = 6s: is that just tangential acceleration? If so, are you supposed to include radial acceleration v^2/r?
2. Your integration is incorrect. As I read it, a is the second derivative of s. Hence, you really have to solve the DE s'' = 6s, with the prescribed initial conditions.

Make sense?
• Sep 29th 2010, 07:03 AM
Ackbeet
Someone pointed out to me that your basic approach in finding the velocity is correct, and you got the correct value for the velocity. However, your tangential acceleration is off by a factor of two. Where did the equation

\$\displaystyle v^{2}=v_{0}^{2}+2a(24)\$

come from? Justify it to me.
• Sep 29th 2010, 03:09 PM
Techro
Looking back i tried another approach.

Assuming acceleration (6s) was the tangential acceleration. Then a = Tangential acceleration + radial acceleration.
Where radial acceleration was v^2/r = 117.901m/s^s. Would that just make tangential acceleration 6s, 6 x 24 ?

And the velocity equation was stupid of me. I tried to solve for acceleration, but I am pretty sure that equation is for when acceleration is constant.
• Sep 29th 2010, 03:41 PM
Ackbeet
Radial and tangential are orthogonal to each other. Think vectors here. In order to find the magnitude of a vector, you do ... what?
• Sep 29th 2010, 03:55 PM
Techro
Use Pythagoras.
square root of \$\displaystyle (an^2 + at^2)\$
where an is the radial acceleration and at is tangential.
• Sep 29th 2010, 04:00 PM
Techro
OMG that was soo stupid. I tried that before and got the wrong answer, I tried it now and got a different answer which worked this time.

Now only 3 more questions to do.
Thank you.
• Sep 29th 2010, 05:10 PM
Ackbeet
You're welcome. Have a good one!