The power exerted by a force on an object is

in which F is the force and v is the velocity of the point of application of the force. Power is a scalar quantity with dimension N-m/s2. Let us illustrate this definition by considering the case of a ball falling from a height h. Both the velocity and the gravitational force are in the same direction, and therefore the power produced must be positive. The velocity of a ball falling a distance d is given by the equation in which v is the speed of the spherical ball and d is the distance of the fall. Thus we obtain the following equation for the power produced by gravity:

in which y is the height of the falling ball at time t and h is the height from which it is falling. Note that the power produced by gravity is not constant but increases with the falling distance. The power exerted by gravity is negative when a ball moves upward in the opposite direction of the gravitational acceleration.

Next consider the power produced by the ground reaction force acting on a spherical ball rolling without slip down an inclined plane (see Fig. 8.2). Because the ball does not slip, the velocity of the point of application of the ground force (point A) is zero. Therefore, the power produced by the ground force is equal to zero.

The power of external forces and couples acting on a rigid body is given by the following equation:

in which the summation is over all forces acting on the body. The terms F and v have their usual meanings. Let us express the velocity of the point of application of F in terms of the velocity of the center of mass and the angular velocity of the body:

v = vc + a X r Using this expression in Eqn. 8.10, we obtain: P = 2(F • (vc + m X r)) P = (2F) • vc + 2(F • a X r) P = (2F) • vc + 2(r X F) • a P = (2F) • vc + (2Mc) • a

in which 2F denotes the resultant force and (2Mc) is the resultant moment about the center of mass. According to Newton's laws of motion, the resultant external force acting on an object is equal to the product of the mass and the acceleration of the center of mass of the object:

where the term Tc represents the kinetic energy associated with the center of mass.

The second term on the right-hand side of Eqn. 8.11 is a product of the resultant moment acting on the rigid object with the angular velocity of the object. Using the principle of conservation of angular momentum, we can express the resultant external moment in terms of the angular acceleration and angular velocity. For the planar motion studied in Chapter 4:

2Mc = Ic a in which Ic is the mass moment of inertia with respect to an axis that passes through the center of mass. The direction of the axis is perpendicular to the plane of motion. The symbol a denotes angular acceleration, as earlier. Substituting this equation on the right-hand side of Eqn. 8.11, we find:

where the symbol Tr represents the kinetic energy of the rigid object that is associated with the rotation around the center of mass. Although we have derived this equation for a rigid body whose plane of motion is parallel to a plane of symmetry of the body, it can be shown that even in the most general three-dimensional motion, mechanical power of external moment acting on a rigid object is equal to the time derivative of the part of the kinetic energy associated with rotation around the center of mass. Thus, we have the principle of conservation of mechanical energy:

According to this equation, the time rate of change of kinetic energy of a rigid body is equal to the rate of work done on it by the external forces and moments. Integrating this equation with respect to time, we arrive at the following relation:

in which T1 and T2 denote the kinetic energy of the body at times t1 and t2 and Wi-2 is the work done on the body by external forces and couples between times t1 and t2.

Example 8.1. Rolling of a Spherical Ball Down an Inclined Plane. A sphere of radius a is released from rest and rolls without sliding down an inclined plane (see Fig. 8.2). Find the velocity of the ball after its center of mass moves a distance b.

Solution: The free-body diagram of the rolling sphere is shown in Fig. 8.2. As noted, the ground reaction force produces no power. Thus, the increase in kinetic energy must be equal to the work done on the ball by the gravitational force:

T2 = (V2) m (v)2 + (V2) (2/5) ma2 (v/a)2 = 0.7 m v2

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