Circular Motion and Gravitation

Uniform circular motion is when objects travel uniformly (constant speed) in a circle.

Uniform circular motion and gravitation feature many of the variables from the kinematics unit, but instead of being tangential (linear), they are now angular.  To convert from angular to tangential values, multiply by the radius of the object or the circle it is travelling in.  Some examples of these variables include ω (angular velocity), α (angular acceleration), θ (angular displacement), etc.  There are a few formulas essential to this unit:

Fc = mv^2/r

ac = v^2/r

t = 1/f

f = 1/t

Fg = Gm1m2/r^2 (Newton’s Law of Universal Gravitation)

g = Gme/r^2

v = √(GMe)/r^2

G is the gravitational constant, which is equal to approximately 6.674×10^−11 m^3⋅kg^−1⋅s^−2

Centripetal force is a concept that will often be tested in most physics classes.  It is a label for the net force acting normal to the object’s motion.  So, centripetal force isn’t actually a force in and of itself, but rather a label for other forces.  The formula for centripetal force is Fc = mac = mv^2/r.  

Example Problem #1

In this image of a tennis ball (.1kg) attached to a rope being swung in a circle (horizontally), it is given that r = 2m.  If the maximum tension the rope can withhold is 20N, what is the maximum tangential velocity that the ball can be swung at?

Answer: 20m/s

Explanation: Since, in this case, Fc = Ft, we can say that Ft = mv^2/r = .1v^2/2.  Since we want to find the velocity at the maximum tension, we would set 20 = .1v^2/2 and solve for v to get 20 m/s.



Gravitation is the tendency of bodies to be attracted to one another.

Gravitation, as previously described, is the force of attraction between objects (like planets, for example).  However large the objects are, there is some force of attraction between them.  Newton’s Law of Universal Gravitation quantifies the gravitational attraction between two bodies.  The formula is Fg = Gm1m2/r^2, where m1 and m2 represent the masses of the two objects and r is the distance between the two objects.

Example Problem #2

What is the gravitational force between the sun and Earth? (mEarth = 5.972 × 10^24 kg, msun = 1.989 × 10^30 kg, and the distance between them is 149.6 billion meters)

Answer: 3.54 x 10^22 N

Explanation: Using the formula Fg = Gm1m2/r^2 and the given values , we can find that Fg = 3.54 x 10^22 N