How does centripetal force depend on mass, radius and speed?

[Aus_Phys]

I had to do an experiment where you manipulate and change different variables (mass providing centripetal force, radius etc.) and then further invesigate the effects on centripetal force that these changes had.
My problem is that it says that by keeping the mass providing the centripetal force constant, the centripetal force remains constant... so what would that force be if the providing mass was 250g. I've been reading and from what I've seen it would be 2.5N but it still doesn't make sense! Can someone please explain!


Also... What would be the relationship between the variables i mentioned?

 

[Aus_Phys]

Plz Help!

 

[gabee]

It sounds like you probably need to review about centripetal force. If an object is moving in a circle at constant tangential speed, there is an acceleration directed toward the center of the circular path. That acceleration (a_c) is v^2 / r (you can prove that geometrically). Since F = m*a, the centripetal force is $F_c = m a_c$, or $F_c = m v^2 / r$. So, centripetal force depends on the mass of the object moving in the circle, its velocity, and the radius of the circular path.

I hope this helps your understanding of what was happening in the experiment. Was the experiment by any chance swinging around a weight on a string through a pipe with another weight attached at the other end?

 

[Aus_Phys]

Yes It Was

 

[Aus_Phys]

Thanks that makes sense, but I'm still not quite sure on the mass providing the force, because it says that by having this mass constant, the force will remain constant, but what i that constant force going to be is the mass is 250g??

 

[gabee]

Alright...the question probably meant that by keeping the mass hanging at the bottom steady while swinging the top mass around, it exerts a constant force (Mg) on the string. The string then exerts that force on the swinging weight (it acts as the centripetal force), and since the string doesn't move up or down in the pipe, the net force is 0; i.e. mv^2/r = Mg, where M is the hanging mass. You probably did several trials in which you varied the swinging weight's mass or the velocity with which you swung it around, but by using the same mass on the bottom you are ensuring that the same force is exerted on the swinging mass every time. Does that help?

 

[Aus_Phys]

Yes, Thank You