Back to chapter

9.15:

Gravitational Potential Energy for Extended Objects

JoVE Core
Physics
A subscription to JoVE is required to view this content.  Sign in or start your free trial.
JoVE Core Physics
Gravitational Potential Energy for Extended Objects

Languages

Share

For a collection of point masses, the position vector for the center of mass is written as the summation of the product of each mass and its position vector divided by the total mass of all the particles.

For extended objects with a uniform mass distribution, the point mass is replaced with the differential mass element and the summation with an integral to obtain the center of mass coordinates.

Consider a free-falling extended object along the y-axis under uniform acceleration due to gravity. The total potential energy of this object is the sum of the potential energy due to each mass element.

Substituting the expression for the center of mass gives the total potential energy as the product of the total mass, the acceleration due to gravity, and the y-coordinate of the center of mass.

This implies that the gravitational potential energy of any extended object can be estimated by considering the object's entire mass to be concentrated at its center of mass.

9.15:

Gravitational Potential Energy for Extended Objects

Consider a system comprising several point masses. The coordinates of the center of mass for this system can be expressed as the summation of the product of each mass and its position vector divided by the total mass:

Equation1

Suppose the point masses are replaced with an extended object with uniformly distributed mass. The coordinates of the center of mass for this object can be obtained by replacing the point mass with the differential mass element and the summation with an integral in the equation for the center of mass:

Equation2

Consider a ring with uniform mass distribution M and radius R. The circular symmetry ensures that the center of mass is located at the ring's geometric center:

Equation1

Consider a coordinate system with its origin located at the center of the ring. Since the ring has a uniform mass distribution, the linear mass density is constant. So, the differential mass element on the surface of the ring is the product of the linear mass density and the differential length element on the ring's surface.

Now, using the expression for the center of mass and substituting the value of the position vector in the component form and the differential mass element gives the following equation:

Equation3

As the arc length ds subtends a differential angle , the arc length equals the radius multiplied by the differential angle. The linear mass density is the total mass divided by the length of the ring. Incorporating these values of arc length and linear mass density, the center of mass expression reduces to the following:

Equation4

The variable of integration is the angle θ. So, the limits of integration around the ring are θ = 0 to θ = 2π. The integral is separated into the x and y components and integrated across the limits:

Equation5

Since the origin of the coordinate is located at the center of the ring, the center of mass of the ring lies at its geometric center.

Suggested Reading

  1. Young, H.D and Freedman, R.A. (2012). University Physics with Modern Physics. San Francisco, CA: Pearson. Pp. 293
  2. OpenStax. (2019). University Physics Vol. 1. [Web version]. Retrieved from https://openstax.org/books/university-physics-volume-1/pages/9-6-center-of-mass