MID-TERM II MOVED TO NOVEMBER 14th

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, … 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 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.

Acceleration of the Center of Mass: Let acm be the acceleration of the cener of mass (the rate of change of vcm with respect to time); then Macm = mAaA + mBaB + ……… The right side of this equation is equal to the vector sum ΣF of all the forces acting on all the particles. We may classify each force as internal or external. The sum of forces on all the particles is then ΣF = ΣFext + ΣFint = Macm

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 –  (radians, rad). Before, most of us thought “in degrees”. Now we must think in radians. Where 1 radian = 57.3o or 2p radians=360o . Try to convert some common angles ( 45o, 90o, 360o).

Unit: rad/s2

MID-TERM II MOVED TO NOVEMBER 14th

Relationship Between Linear and Angular Quantities

v = rω atan = rα arad = rω2

Kinetic Energy of Rotating Rigid Body Moment of Inertia KA = (1/2)mAvA2 vA = rA ω vA2 = rA2 ω2 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

Rotational energy

Moments of inertia & rotation

Rotation about a Moving Axis Every motion of 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 asociated with rotation about an axis through the center of mass

Ktotal = (1/2)Mvcm2 + (1/2)Icmω2 Total Kinetic Energy Ktotal = (1/2)Mvcm2 + (1/2)Icmω2

A rotation while the axis moves

Race of the objects on a ramp