Show that the total kinetic energy of a system of point masses, which equals ½ m

Question

Show that the total kinetic energy of a system of point masses, which equals ½ mi vi middot vi, can be written as the sum of the kinetic energy of the center of mass, i.e., ½ M total v2cm, plus the kinetic energy of the point masses about the center of mass, i.e., ½ mi(v i)2. Note that M total is M total = mi. Consider three masses, m1, m2, and m3, such that m1 = 3m, m2 = 2m, and m3 = lm. Let the locations and velocities of m1, m2, and m3 be x1 = 3 e^x + 4 e^y, x2 = - 1 e^x + - 1 e^y, x3 = - 1 e^x + 5 e^y (in units of meters) and v1 = 4 e^x + 4 e^y, v2 = - 3 e^y, v3 = - 2 e^x (in units of meters per second). What are the center of mass position and the center of mass velocity? What are x 1, x 2, and x 3? What are v 1, v 2, and v 3? Show that v i = 0 for these three masses. Show that the kinetic energy of the center of mass plus the kinetic energy about the center of mass equals the total kinetic energy of the three masses. Draw a picture representing x1, x2, x3, x 1, x 2, x 3, v1, v2, v3, the center of mass position and the center of mass velocity.

Explanation / Answer

given Mtotal = mi

we can say

velocity of point mass about the center of mass is the difference between velocity of point mass and velocity of center of mass i.e,

(vi')2= vi2-vcm2

so vi2 = vi'2 + vcm2

multiply both sides by 1/2 * m

1/2 m vi^2 =( vi'^2 +vcm^2)1/2 m

so for all point masses

1/2 m vi^2 = ( vi'^2 +vcm^2)1/2 m

1/2 m vi^2 = 1/2 Mtotal vcm^2 + 1/2 mi vi'^2

LHS = total kinetic energy of system of masses

RHS = kinetic energy of center of mass + kinetic energy of point masses about the centre of mass

Hence proved.

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.