# Force acting on an object

1. Mar 10, 2016

### Guessit

1. The problem statement, all variables and given/known data

http://imgur.com/WeZ2lSS

If the image isn't clear, this is what it states:

The speed of a particle of mass m increases at a constant rate as it moves along the path shown from location 1 to 2 and so on. The particle's speed is much less than c, the speed of light, throughout its motion.

Problem:

At which particle location/locations is the magnitude of the transverse component of the time rate of change of momentum ( mag. of dp / dt, perpendicular) the greatest?

At which location is the magnitude of the net force acting on the particle smallest?

2. Relevant equations

F = dp/dt and F = mv^2 / r (I think)

3. The attempt at a solution

This question just stumped me during the exam. If the particle was speeding up at a constant rate, there was a constant force applied to it, but I do not understand how it goes tangential after a point and then makes a path with a smaller radius. Also, looking at the figure, since r was small and v was large at positions 3 and 4, I chose option E (3 and 4) for the first part since F perpendicular seemed the greatest there by mv^2/r. I also thought that since the path followed looks circular, the force experienced at 3 and 4 was the same. I do not understand how it could be circular if the speed at 4 was greater than the speed at 3.

For the second part, I picked E (same at all points) since the speed was increasing at a constant rate and I figured this implied constant acceleration.

2. Mar 10, 2016

### haruspex

No. All that tells you is that the tangential component of the acceleration is constant in magnitude. The normal component of acceleration does not affect speed.

3. Mar 10, 2016

### Guessit

Ahh, that makes a lot more sense. So for problem 1, since they are asking about the normal component of force, would that be the same at points 3 and 4 since the path is circular and the normal component of force would be the same at both points?

And for problem 2, the least force would be at point 2 then since there is only a tangential component for the acceleration whereas at 1,3 and 4, there is also a normal component, and hence net force would be the vector sum of those 2. Is that right?

4. Mar 10, 2016

### SammyS

Staff Emeritus
compare the centripetal force at 3 with that at 4.

5. Mar 10, 2016

### Guessit

mv^2/r ... and v increases, which means there is a greater centripetal force at 4. Whoops! Looks like I messed up on this one :/

Thanks guys!