(4.1)
(4.2)
Linear momentum is a vector quantity because it equals the product of a scalar quantity m and a vector quantity v. Its direction is along v, it has dimensions ML/T, and its SI unit is kg · m/s.
If a particle is moving in an arbitrary direction, p must have three components
(4.3)
As m=const:
The time rate of change of the linear momentum of a particle is equal to the net force acting on the particle.
(4.5)
(4.4)
Whenever two or more particles in an isolated system interact, the total momentum of the system remains constant.
This law tells us that the total momentum of an isolated system at all times equals its initial momentum.
Notice that we have made no statement concerning the nature of the forces acting on the particles of the system. The only requirement is that the forces must be internal to the system.
(4.6)
Equation 4.8 is an important statement known as the impulse–momentum theorem:
The impulse of the force F acting on a particle equals the change in the momentum of the particle.
Figure 4.2 (a) A force acting on a particle may vary in time. The impulse imparted to the particle by the force is the area under the force-versus-time curve. (b) In the time interval (t, the time-averaged force (horizontal dashed line) gives the same impulse to a particle as does the time-varying force described in part (a).
(a)
(b)
(4.10)
(4.12)
The calculation becomes especially simple if the force acting on the particle is constant. In this case, and Equation 4.11 becomes
In many physical situations, we shall use what is called the impulse approximation, in which we assume that one of the forces exerted on a particle acts for a short time but is much greater than any other force present.
(4.11)
Figure 4.3 (a) The collision between two objects as the result of direct contact. (b) The “collision” between two charged particles.
That is, the particles exchange velocities if they have equal masses. This is approximately what one observes in head-on billiard ball collisions - the cue ball stops, and the struck ball moves away from the collision with the same velocity that the cue ball had.
If m2 is much greater than m1 and particle 2 is initially at rest, then v1f ≈ -v1i and v2f ≈ 0.
For such two-dimensional collisions, we obtain two component equations for conservation of momentum:
If the collision is inelastic, kinetic energy is not conserved and Equation 4.26 does not apply.
(4.27)
(4.28)
(4.29)
The center of mass of any symmetric object lies on an axis of symmetry and on any plane of symmetry.
a continuous mass distribution
where vi is the velocity of the ith particle. Rearranging Equation 4.34 gives
(4.34)
(4.35)
Therefore, we conclude that the total linear momentum of the system equals the total mass multiplied by the velocity of the center of mass. In other words, the total linear momentum of the system is equal to that of a single particle of mass M moving with a velocity vCM.
Rearranging this expression and using Newton’s second law, we obtain
where Fi is the net force on particle i.
(4.36)
(4.37)
The center of mass of a system of particles of combined mass M moves like an equivalent particle of mass M would move under the influence of the net external force on the system.
(4.38)
Figure 5.1 A compact disc rotating about a fixed axis through O perpendicular to the plane of the figure.
Rotation of a Rigid Object About a Fixed Axis
(5.1)
The average angular speed
The instantaneous angular speed
(5.2)
(5.3)
(5.4)
(5.5)
If we eliminate α between Equations 5.6 and 5.7, we obtain
(5.8)
(5.9)
(5.10)
That is, the tangential speed of a point on a rotating rigid object equals the perpendicular distance of that point from the axis of rotation multiplied by the angular speed.
That is, the tangential component of the linear acceleration of a point on a rotating rigid object equals the point’s distance from the axis of rotation multiplied by the angular acceleration.
(5.11)
A point moving in a circular path undergoes a radial acceleration ar of magnitude v2/r directed toward the center of rotation .
Because v = rω for a point P on a rotating object, we can express the centripetal acceleration at that point in terms of angular speed as
(5.12)
(5.13)
(5.15)
Where KR is Rotational kinetic energy.
(5.16)
When a force is exerted on a rigid object pivoted about an axis, the object tends to rotate about that axis. The tendency of a force to rotate an object about some axis is measured by a vector quantity called torque τ (Greek tau).
(5.18)
(5.20)
So, again we see that the net torque about the rotation axis is proportional to the angular acceleration of the object, with the proportionality factor being I, a quantity that depends upon the axis of rotation and upon the size and shape of the object.
(5.21)
The work done by F on the object as it rotates through an infinitesimal distance
where Fsinφ is the tangential component of F, or, in other words, the component of the force along the displacement.
(5.22)
(5.23)
(5.25)
(5.26)
Quick Quiz 2 A car and a large truck traveling at the same speed make a head-on collision and stick together. Which vehicle experiences the larger change in the magnitude of momentum? (a) the car (b) the truck (c) The change in the magnitude of momentum is the same for both. (d) impossible to determine.
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