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The explanation of this apparently astonishing coincidence is: We defined the center of mass precisely so this is exactly what we would get. Recall that first we defined the momentum of the system:

p CM = j = 1 N d p j d t .

We then concluded that the net external force on the system (if any) changed this momentum:

F = d p CM d t

and then—and here’s the point—we defined an acceleration that would obey Newton’s second law. That is, we demanded that we should be able to write

a = F M

which requires that

a = d 2 d t 2 ( 1 M j = 1 N m j r j ) .

where the quantity inside the parentheses is the center of mass of our system. So, it’s not astonishing that the center of mass obeys Newton’s second law; we defined it so that it would.

Summary

  • An extended object (made up of many objects) has a defined position vector called the center of mass.
  • The center of mass can be thought of, loosely, as the average location of the total mass of the object.
  • The center of mass of an object traces out the trajectory dictated by Newton’s second law, due to the net external force.
  • The internal forces within an extended object cannot alter the momentum of the extended object as a whole.

Conceptual questions

Suppose a fireworks shell explodes, breaking into three large pieces for which air resistance is negligible. How does the explosion affect the motion of the center of mass? How would it be affected if the pieces experienced significantly more air resistance than the intact shell?

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Problems

Three point masses are placed at the corners of a triangle as shown in the figure below.

A right triangle with sides length 3 c m and 4 c m has masses of 100 g at the vertex between the hypotenuse and the 4 c m side, 75 g at the vertex between the hypotenuse and the 3 c m side, and 150 g at the vertex between the the 3 c m side and the 4 c m side.

Find the center of mass of the three-mass system.

With the origin defined to be at the position of the 150-g mass, x CM = −1.23 cm and y CM = 0.69 cm

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Two particles of masses m 1 and m 2 separated by a horizontal distance D are released from the same height h at the same time. Find the vertical position of the center of mass of these two particles at a time before the two particles strike the ground. Assume no air resistance.

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Two particles of masses m 1 and m 2 separated by a horizontal distance D are let go from the same height h at different times. Particle 1 starts at t = 0 , and particle 2 is let go at t = T . Find the vertical position of the center of mass at a time before the first particle strikes the ground. Assume no air resistance.

y CM = { h 2 1 4 g t 2 , t < T h 1 2 g t 2 1 4 g T 2 + 1 2 g t T , t T

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Two particles of masses m 1 and m 2 move uniformly in different circles of radii R 1 and R 2 about origin in the x , y -plane. The x - and y -coordinates of the center of mass and that of particle 1 are given as follows (where length is in meters and t in seconds):
x 1 ( t ) = 4 cos ( 2 t ) , y 1 ( t ) = 4 sin ( 2 t )

and:
x CM ( t ) = 3 cos ( 2 t ) , y CM ( t ) = 3 sin ( 2 t ) .

  1. Find the radius of the circle in which particle 1 moves.
  2. Find the x - and y -coordinates of particle 2 and the radius of the circle this particle moves.
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Two particles of masses m 1 and m 2 move uniformly in different circles of radii R 1 and R 2 about the origin in the x , y -plane. The coordinates of the two particles in meters are given as follows ( z = 0 for both). Here t is in seconds:
x 1 ( t ) = 4 cos ( 2 t ) y 1 ( t ) = 4 sin ( 2 t ) x 2 ( t ) = 2 cos ( 3 t π 2 ) y 2 ( t ) = 2 sin ( 3 t π 2 )

  1. Find the radii of the circles of motion of both particles.
  2. Find the x - and y -coordinates of the center of mass.
  3. Decide if the center of mass moves in a circle by plotting its trajectory.

a. R 1 = 4 m , R 2 = 2 m ; b. X CM = m 1 x 1 + m 2 x 2 m 1 + m 2 , Y CM = m 1 y 1 + m 2 y 2 m 1 + m 2 ; c. yes, with R = 1 m 1 + m 2 16 m 1 2 + 4 m 2 2

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Find the center of mass of a one-meter long rod, made of 50 cm of iron (density 8 g cm 3 ) and 50 cm of aluminum (density 2.7 g cm 3 ).

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Find the center of mass of a rod of length L whose mass density changes from one end to the other quadratically. That is, if the rod is laid out along the x -axis with one end at the origin and the other end at x = L , the density is given by ρ ( x ) = ρ 0 + ( ρ 1 ρ 0 ) ( x L ) 2 , where ρ 0 and ρ 1 are constant values.

x c m = 3 4 L ( ρ 1 + ρ 0 ρ 1 + 2 ρ 0 )

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Find the center of mass of a rectangular block of length a and width b that has a nonuniform density such that when the rectangle is placed in the x , y -plane with one corner at the origin and the block placed in the first quadrant with the two edges along the x - and y -axes, the density is given by ρ ( x , y ) = ρ 0 x , where ρ 0 is a constant.

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Find the center of mass of a rectangular material of length a and width b made up of a material of nonuniform density. The density is such that when the rectangle is placed in the xy -plane, the density is given by ρ ( x , y ) = ρ 0 x y .

( 2 a 3 , 2 b 3 )

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A cube of side a is cut out of another cube of side b as shown in the figure below.

A large cube of side b has a cube of side a cut out of its bottom left front corner.

Find the location of the center of mass of the structure. ( Hint: Think of the missing part as a negative mass overlapping a positive mass.)

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Find the center of mass of cone of uniform density that has a radius R at the base, height h , and mass M . Let the origin be at the center of the base of the cone and have + z going through the cone vertex.

( x CM , y CM , z CM ) = ( 0,0 , h / 4 )

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Find the center of mass of a thin wire of mass m and length L bent in a semicircular shape. Let the origin be at the center of the semicircle and have the wire arc from the + x axis, cross the + y axis, and terminate at the − x axis.

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Find the center of mass of a uniform thin semicircular plate of radius R . Let the origin be at the center of the semicircle, the plate arc from the + x axis to the −x axis, and the z axis be perpendicular to the plate.

( x CM , y CM , z CM ) = ( 0 , 4 R / ( 3 π ) , 0 )

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Find the center of mass of a sphere of mass M and radius R and a cylinder of mass m , radius r , and height h arranged as shown below.

Figure a has a sphere on top of a vertical cylinder. Figure b has a sphere centered on top of a horizontal cylinder.

Express your answers in a coordinate system that has the origin at the center of the cylinder.

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Questions & Answers

V=E½-P-½ where v; velocity, P; density and E; constant. Find dimension and it's units of E (constant)
michael Reply
ML-3
LAWAL
derivation of simple harmonic equation
Daud Reply
if an equation is dimensionally correct does this mean that equation must be true?
michael Reply
how do I calculate angular velocity
Priscilla Reply
w=vr where w, angular velocity. v; velocity and r; radius of a circle
michael
sorry I meant Maximum positive angular velocity of
Priscilla
please can u tell me the formular for tension in angular velocity I kind of forget it please don't ignore this msg I need it nw
sodeeq
Can any one give me the definition for Bending moment plz...
Prema Reply
I need a question for moment
paul Reply
what is charge
Zarshad
An attribution of particle that we have thought about to explain certain things like Electomagnetism
Nikunj
please what is the formula instantaneous velocity in projectile motion
Isaiah Reply
A computer is reading from a CD-ROM that rotates at 780 revolutions per minute.What is the centripetal acceleration at a point that is 0.030m from the center of the disc?
Rapqueen Reply
change revolution per minute by multiplying from 2pie and devide by 60.and take r=.030 and use formula centripital acceleration =omega sqare r.
Kumar
OK thank you
Rapqueen
observation of body boulded
Anwer Reply
a gas is compressed to 1/10 0f its original volume.calculate the rise temperature if the original volume is 400k. gamma =1.4
Celine Reply
the specific heat of hydrogen at constant pressure and temperature is 14.16kj|k.if 0.8kg of hydrogen is heated from 55 degree Celsius to 80 degree Celsius of a constant pressure. find the external work done .
Celine
hi
shaik
hy
Prasanna
g
Ahmad
what is imaginary mass and how we express is
Yash Reply
what is imaginary mass how we express it
Yash
centre of mass is also called as imaginary mass
Lokmani
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Khlil Reply
Many amusement parks have rides that make vertical loops like the one shown below. For safety, the cars are attached to the rails in such a way that they cannot fall off. If the car goes over the top at just the right speed, gravity alone will supply the centripetal force. What other force acts and what is its direction if: (a) The car goes over the top at faster than this speed? (b) The car goes over the top at slower than this speed?
English Reply
how can I convert mile to meter per hour
Folorunsho Reply
1 mile * 1609m
Boon
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Shrushti Reply
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Practice Key Terms 4

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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