The mathematics of circular motion.

[ajassat]

I am a high school student with a general interest in physics of all types.

We have learned that circular motion is possible because of the centripetal force acting towards the centre. I gather that the centripetal force is the resultant of the gravitational and outward forces. The object moving around the circle is accelerating because the direction of the object changes at tangents to the circles.

I would like to know about the mathematics which proves this. Can someone introduce some equations and explain them to me?

Where is the math which governs the net centripetal force?
Can we calculate the angular momentum?

Regards,
Adam

 

[jtbell]

I gather that the centripetal force is the resultant of the gravitational and outward forces.

The centripetal force is simply the resultant of all forces acting on the object. In most situations where we discuss circular motion, there is no outward force acting on the object. For example, the only (significant) force on a satellite orbiting the Earth is the (inward) gravitational force exerted by the Earth.

I would like to know about the mathematics which proves this. Can someone introduce some equations and explain them to me?

Perhaps this page might help:

http://hyperphysics.phy-astr.gsu.edu/Hbase/cf.html

 

[ajassat]

The centripetal force is simply the resultant of all forces acting on the object. In most situations where we discuss circular motion, there is no outward force acting on the object. For example, the only (significant) force on a satellite orbiting the Earth is the (inward) gravitational force exerted by the Earth.



Perhaps this page might help:

http://hyperphysics.phy-astr.gsu.edu/Hbase/cf.html

The link was good in the sense it told me the formula and I was able to practice some calculations. However, using which logic and reasoning do we come to that formula?

 

[rcgldr]

Covered, but not explained well on that same web page:

http://hyperphysics.phy-astr.gsu.edu/Hbase/cf.html#cf2

given
1. S/r = dV/V
2. S = V dt

Solve 1. for S

S = r dv/V

substitute in 2.

r dv/V = V dt

rearange this

dv/dt = V^2 / r

a = dv/dt = V^2 / r

 

[ajassat]

Covered, but not explained well on that same web page:

http://hyperphysics.phy-astr.gsu.edu/Hbase/cf.html#cf2

given
1. S/r = dV/V
2. S = V dt

Solve 1. for S

S = r dv/V

substitute in 2.

r dv/V = V dt

rearange this

dv/dt = V^2 / r

a = dv/dt = V^2 / r

Excellent. I now see the logic behind the formula. Thank you.