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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. Center of Mass for Two Objects X cm = (m 1 x 1 + m 2 x 2 )/(m 1 + m 2 ) = (m 1 x 1 + m 2 x 2 )/M. Locating the Center of Mass.
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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.
Center of Mass for Two Objects Xcm = (m1x1 + m2x2)/(m1 + m2) = (m1x1 + m2x2)/M
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.
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 + ………).
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 + ………).
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.
CHAPTER 9 ROTATIONAL MOTION
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.
Angular displacement : Δθ (radians, rad). • Before, most of us thought “in degrees”. • Now we must think in radians. Where 1 radian = 57.3o or 2p radians=360o .
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
An electric fan is turned off, and its angular velocity decreases • uniformly from 500 rev/min to 200 rev/min in 4.00 s . • 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?
v = rω atan = rα arad = rω2
Kinetic Energy of Rotating Rigid BodyMoment of Inertia KA = (1/2)mAvA2 vA = rAω vA2 = rA2ω2 KA = (1/2)(mArA2)ω2 KB = (1/2)(mBrB2)ω2 KC = (1/2)(mCrC2)ω2 .. K = KA + KB + KC + KD …. K = (1/2)(mArA2)ω2 + (1/2)(mBrB2)ω2 ….. K = (1/2)[(mArA2) + (mBrB2)+ …] ω2 K = (1/2) I ω2 I = mArA2 + mBrB2 + mCrC2) + mDrD2 + … Unit: kg.m2 A rA B rB C rC
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
Total Kinetic Energy Ktotal = (1/2)Mvcm2 + (1/2)Icmω2