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The center of mass of a system of masses is the point where the system can be balanced in a uniform gravitational field.

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Page 1: The center of mass of a system of masses is the point where the system can be balanced in a uniform gravitational field

The center of mass of a system of

masses is the point where the system

can be balanced in a uniform gravitational

field.

Page 2: The center of mass of a system of masses is the point where the system can be balanced in a uniform gravitational field

Center of Mass for Two ObjectsXcm = (m1x1 + m2x2)/(m1 + m2) = (m1x1 + m2x2)/M

Page 3: The center of mass of a system of masses is the point where the system can be balanced in a uniform gravitational field
Page 4: The center of mass of a system of masses is the point where the system can be balanced in a uniform gravitational field

Locating the Center of Mass

In an object of continuous,

uniform mass distribution, the center of

mass is located at the

geometric center of the

object. In some cases, this means

that the center of mass is not located within

the object.

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Page 6: The center of mass of a system of masses is the point where the system can be balanced in a uniform gravitational field

Suppose we have several particles A, B, etc., with masses mA, mB, …. Let the coordinates of A be (xA, yA), let those of B be (xB, yB), and so on. We define the center of mass of the system as the point having coordinates (xcm,ycm) given by

xcm = (mAxA + mBxB + ……….)/(mA + mB + ………),

Ycm = (mAyA + mByB +……….)/(mA + mB + ………).

Page 7: The center of mass of a system of masses is the point where the system can be balanced in a uniform gravitational field

a=1m

Page 8: The center of mass of a system of masses is the point where the system can be balanced in a uniform gravitational field

The velocity vcm of the center of mass of a collection of particles is the mass-weighed average of the velocities of the individual particles:

vcm = (mAvA + mBvB + ……….)/(mA + mB + ………).

In terms of components,

vcm,x = (mAvA,x + mBvB,x + ……….)/(mA + mB + ………),

vcm,y = (mAvA,y + mBvB,y + ……….)/(mA + mB + ………).

Page 9: The center of mass of a system of masses is the point where the system can be balanced in a uniform gravitational field

For a system of particles, the momentum P of the center of mass is the total mass M = mA + mB +…… times the velocity vcm of the center of mass:

Mvcm = mAvA + mBvB + ……… = P

It follows that, for an isolated system, in which the total momentum is constant the velocity of the center of mass is also constant.

Page 10: The center of mass of a system of masses is the point where the system can be balanced in a uniform gravitational field

CHAPTER 9ROTATIONAL MOTION

Page 11: The center of mass of a system of masses is the point where the system can be balanced in a uniform gravitational field

Goals for Chapter 9

• To study angular velocity and angular acceleration.

• To examine rotation with constant angular acceleration.

• To understand the relationship between linear and angular quantities.

• To determine the kinetic energy of rotation and the moment of inertia.

• To study rotation about a moving axis.

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Page 13: The center of mass of a system of masses is the point where the system can be balanced in a uniform gravitational field
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Page 15: The center of mass of a system of masses is the point where the system can be balanced in a uniform gravitational field

•Angular displacement : Δθ (radians, rad).– Before, most of us thought “in degrees”.– Now we must think in radians. Where 1 radian = 57.3o

or 2radians=360o .

Page 16: The center of mass of a system of masses is the point where the system can be balanced in a uniform gravitational field
Page 17: The center of mass of a system of masses is the point where the system can be balanced in a uniform gravitational field
Page 18: The center of mass of a system of masses is the point where the system can be balanced in a uniform gravitational field
Page 19: The center of mass of a system of masses is the point where the system can be balanced in a uniform gravitational field

Unit: rad/s2

Page 20: The center of mass of a system of masses is the point where the system can be balanced in a uniform gravitational field

Comparison of linear and angular For linear motion with constant acceleration

For a fixed axis rotation with constant angular acceleration

a = constant

v = vo + at

x = xo + vot + ½at2

v2 = vo2 + 2a(x-xo)

x–xo = ½(v+vo)t

α = constant

ω = ωo + αt

Θ = Θo + ωot + ½ αt2

ω2 = ωo2 + 2α(Θ-Θo)

Θ-Θo= ½(ω+ωo)t

Page 21: The center of mass of a system of masses is the point where the system can be balanced in a uniform gravitational field
Page 22: The center of mass of a system of masses is the point where the system can be balanced in a uniform gravitational field

An electric fan is turned off, and its angular velocity decreases uniformly from 500  rev/min to 200 rev/min in 4.00 s .

a) Find the angular acceleration in rev/s2 and the number of revolutions made by the motor in the 4.00 s interval.

b) The number of revolutions made in 4.00n s

c) How many more seconds are required for the fan to come to rest if the angular acceleration remains constant at the value calculated in part A?

Page 23: The center of mass of a system of masses is the point where the system can be balanced in a uniform gravitational field

Relationship Between Linear and Angular Quantities

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Page 25: The center of mass of a system of masses is the point where the system can be balanced in a uniform gravitational field

v = rω

atan = rα

arad = rω2

22tan radaaa

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Page 28: The center of mass of a system of masses is the point where the system can be balanced in a uniform gravitational field

Kinetic Energy and Moment of Inertia

Page 29: The center of mass of a system of masses is the point where the system can be balanced in a uniform gravitational field

Kinetic Energy of Rotating Rigid BodyMoment of Inertia

KA = (1/2)mAvA2

vA = rA ω vA2 = rA

2 ω2 KA = (1/2)(mArA

2)ω2

KB = (1/2)(mBrB2)ω2

KC = (1/2)(mCrC2)ω2

..

K = KA + KB + KC + KD ….

K = (1/2)(mArA2)ω2 + (1/2)(mBrB

2)ω2 …..K = (1/2)[(mArA

2) + (mBrB2)+ …] ω2

K = (1/2) I ω2 I = mArA

2 + mBrB2 + mCrC

2) + mDrD2 + …

Unit: kg.m2

A

B

C

rA

rB

rC

Page 30: The center of mass of a system of masses is the point where the system can be balanced in a uniform gravitational field

Rotational energy

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Page 32: The center of mass of a system of masses is the point where the system can be balanced in a uniform gravitational field

Moments of inertia & rotation

Page 33: The center of mass of a system of masses is the point where the system can be balanced in a uniform gravitational field

Rotation about a Moving Axis

• Every motion of a rigid body can be represented as a combination of motion of the center of mass

(translation) and rotation about an axis through the center of mass

• The total kinetic energy can always be represented as the sum of a part associated with motion of the center of mass (treated as a point) plus a part associated with rotation about an axis through the center of mass

Page 34: The center of mass of a system of masses is the point where the system can be balanced in a uniform gravitational field

Total Kinetic Energy

Ktotal = (1/2)Mvcm2 + (1/2)Icmω2

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Race of the objects on a ramp •

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A rotation while the axis moves