Newtons Laws of Motion and Their Applications

We have explored the physical content of the laws of motion in the introduction of this chapter. The mathematical operations presented in the preceding sections now allow us to write these laws in mathematical language. According to Newton's first law, the resultant force acting on a particle must be equal to zero when the particle is at rest or moving with constant velocity in an inertial reference frame:

in which 2F denotes the sum of all forces acting on the particle. Newton's second law relates the resultant force to the acceleration of the particle

in which m is the mass of the particle and a is its acceleration with respect to an inertial reference frame.

According to the third law, the force of action is equal in magnitude and opposite in direction to the force of reaction f1-2 = —2-1 (2.11c)

in which f1-2 represents the force exerted on particle 1 by particle 2.

The term particle is used in the description of these mathematical formulations to represent an object, small or large, with the stipulation that the variation of velocity (acceleration) within the object is negligible compared to the mean velocity (acceleration) of the object. In his studies, Newton considered earth and the stars as particles. When considering the trajectory of a football in air, it is appropriate to consider the ball as a particle. However, when studying the spin of the ball as it traverses the air, the size and the shape of the ball must be taken into account.

Example 2.5. Free Fall of an Object: An Experiment by Galileo. The legend is that Galileo determined the law of gravity by conducting an ex periment in which iron balls of different sizes were dropped from the top of the Pisa tower and a group of his friends at different floors of the tower measured the time of the free fall at specified heights. Iron balls of varying diameter fall toward earth with constant acceleration (Fig. 2.8). This acceleration is called gravitational acceleration. Its magnitude is usually denoted by g. The force that causes gravitational acceleration is the gravitational force (the weight of the object).

With respect to a Cartesian coordinate system fixed on earth, the position of a free-falling iron ball varies with time in accordance with the following equation:

r = 0 ei + (ho - gt2/2) e2 + 0 = (ho - gt2/2) e2 (2.12a)

where x1, x2, and x3 are the projections of r along the coordinate axes, ho is the height of the Pisa tower (~80 m), and t denotes time in seconds (Fig. 2.8).

Taking the time derivative of Eqn. 2.12, one can compute the velocity and the acceleration of an iron ball falling under the action of gravity:

According to Eqn. 2.13, the speed of an iron ball falling in air (||v||) increases with time whereas its acceleration remains constant. Furthermore, acceleration is also not dependent on the mass of the ball.

Example 2.6. Terminal Velocity of a Snowflake. When a body moves in a medium such as air or water, the medium exerts a resistance force on

mg mg mg mg

Figure 2.8. A steel ball dropped from the Pisa tower. The black circles indicate the position of a steel ball at equal time intervals. The only force acting on the ball is the force of gravity.

e the body. The resistance force increases with the velocity of the motion relative to the fluid. It is also dependent on the shape of the moving body and on its orientation relative to the direction of motion. Resistance force depends nonlinearly on the velocity and geometry of the body. However, for very small relative velocities (to 1 m/s), resistance force is proportional to the first power of velocity. Consider the case of a snowflake falling under gravity and retarded by an air resistance. Let u be the downward velocity of the snowflake and let (—k u) be the frictional resistance per unit mass of the snowflake, k being a constant resistance coefficient. Determine the trajectory of the snowflake. What is its terminal velocity?

Solution: In one-dimensional motion where velocity, force, and acceleration all point in the same direction, it is not necessary to use the vector notation. The equation of motion for the snowflake in the direction of gravity can be written as follows:

where (du/dt) is the acceleration of the snowflake, considered positive downward. This differential equation can be integrated analytically with respect to time to yield u = (dx/dt) = g/k — (C/k) e—k t (2.15a)

where x is the position of the snowflake, measured downward from a reference point in air (dx/dt = u), and C is an arbitrary constant to be determined by the initial conditions at t = 0. The exponential term in Eqn. 2.15 rapidly decreases toward zero with increasing time. Thus, the speed u of the falling snowflake becomes (g/k) for sufficiently large times. This constant velocity is termed the terminal velocity of the snowflake. Suppose the terminal velocity of a snowflake is 1.2 m/s; then the resistance coefficient k must be equal to 8.175 s—1. For this k value the term e—kt is practically equal to zero when t = 1 s. The velocity of the snowflake approaches its terminal value within a fraction of a second. Experiments indicate that the resistance coefficient k increases with the mass density of air. This finding is consistent with the common observation that the snowflakes fall with smaller velocities in colder climates.

Example 2.7. Motion of Spherical Balls on an Inclined Plane. In a freefall experiment, a heavy body rapidly gains speed so that it becomes difficult to record the falling distance as a function of time. To bypass this difficulty, Galileo designed an experiment in which the iron balls were set free on top of a smooth plane that made an angle a with the horizontal plane (Fig. 2.9). The iron balls slid down the inclined plane much slower than the velocities observed during a free fall.

Two distinct forces act on an iron ball moving down an inclined plane: the gravitational force mg pulling the ball downward and the force N ex-

Figure 2.9. A steel ball sliding down a smooth (frictionless) plane. The diagram shows the forces acting on the ball as it speeds down the inclined plane.

e erted by the smooth plane on the steel ball. This latter force is called the contact force. Its function is to keep the ball from cutting into the inclined platform. Galileo polished the surface of the inclined plane as well as that of the spherical balls to reduce the frictional resistance to motion. Therefore, the contact force N acted in the direction normal to the plane of motion as shown in Fig. 2.9. Let us denote the magnitude of this contact force by N. Newton's second law for the iron balls moving down an inclined plane can be written as mg sin a e1 - mg cos a e2 + N e2 = m a1 e1 (2.16a)

This vectorial equality is equivalent to the following two scalar equations:

According to Eqn. 2.16b, acceleration of an iron ball sliding down an inclined plane is sin a times smaller than the acceleration during free fall. For example, if a is chosen as 5°, then sin a = 0.087, and the acceleration of the sliding ball would be equal to 0.85 m/s2 as opposed to 9.81 m/s2 for the free fall.

Integrating acceleration a1 with respect to time, we find the following expressions for velocity (o1) and distance (s1) traveled:

in which v10 and s10 are arbitrary constants to be determined by initial conditions. The velocity of an iron ball is zero at the instant it is released from rest (t = 0) and thus v10 = 0. The parameter s10 can be set equal to zero if the distance s is measured from the point where the ball is released from rest. Thus:

Example 2.8. Oscillation of Pendulum. Galileo understood that to discover the laws of motion, one would have to be able to measure time with better accuracy than permitted by an hourglass. He used his pulse to observe that the period of oscillation of a given pendulum was approximately constant. His interest in the properties of pendulums arose while looking at a chandelier that swung from the roof of the baptistry in Pisa. He attached a bob to the end of a string and swung it. He determined that the period of the pendulum, the time it takes to swing from extreme right to extreme left and back, did not change with the weight of the bob. The maximum angle of swing only had a minor effect on the period of the pendulum. On the other hand, the period varied in proportion with the square root of the length of the pendulum. Galileo used pendulums to measure time in his famous experiments on mechanics.

An equation for the period of a pendulum can be obtained by using Newton's equations of motion. Consider a pendulum with length L and mass m (Fig. 2.10). The forces acting on the bob are the weight mg and the tension T in the string. Newton's second law in the direction of the string (shown by the unit vector er) and in the direction normal to the string (shown by the unit vector et) result in the following equations:

where Q is the angle the pendulum makes with the vertical axis and T is the magnitude of the tension (pulling force) applied by the string on the

Figure 2.10. A pendulum of mass m and length L. The symbol 6 denotes the angle the pendulum makes with the horizontal axis; T is the tension applied by the thin rod on the bob.

mass m. The acceleration of the bob was determined by using Eqn. 2.9c, specific to circular paths.

There are two unknowns in Eqn. 2.17a: the tension T and the angle of swing 6. So, we begin with Eqn. 2.17b in which time t is the independent variable and 6 is the only dependent variable. This is a second-order ordinary differential equation. For this differential equation to have a unique solution, a set of initial conditions must be satisfied. As initial conditions, we specify that the mass m is held at its maximum elevation (6 = 6*) and is let go at t = 0 with zero velocity. In mathematical language, the initial conditions are

where 6* is the maximum angle of swing.

Equation 2.17b can be solved analytically by noting the following equality:

(d26/dt2) = d(d6/dt)/dt = [d(d6/dt)/d6] (d6/dt) (2.17d)

Substituting this equation into Eqn. 2.17b and integrating with respect to 6 one obtains:

Subsequent integration of Eqn. 2.17e with respect to time involves an elliptic integral. The result can be found in some integral tables. For 6* = w/2, the period of the pendulum (the time it takes for the pendulum to complete a whole swing and reach the same spatial point) can be shown to be equal to t* = 7.45 (L/g)1/2 (2.18a)

For small angles of swing, Eqn. 2.17b governing the swing of the pendulum can be simplified by using the approximation sin 6 = 6:

This is a second-order linear and homogeneous differential equation. Its solution can be written as

where A and B are arbitrary constants to be determined by initial conditions. For 6 = 6* and (d6/dt) = 0 at t = 0:

The period t* of this equation can be found by dividing 2w by (g/L)1/2:

Note that this period is independent of the amplitude of oscillation (6*). Comparison of Eqns. 2.18a and 2.18e shows that when the angle of swing is increased from 0 to n/2, the period increases by only 18%. That is why, when we observe children swing, that the period of the swing does not seem to change with the maximum angle of swing.

Example 2.9. Trajectory of a Golf Ball. A man hits a golf ball with initial speed Vo and at an angle of a from the horizontal plane as shown in Fig. 2.11. Determine the trajectory of the ball.

Solution: Neglecting air friction, the only force acting on the ball once it is off the ground is the gravitational force mg. With respect to the coordinate system E that is fixed on earth, the equation of motion reduces to

-mg e2 = m (dv1/dt e1 + dv2/dt e2 +dv3/dt e3) This is, in effect, three scalar equations:

Getting Started With Dumbbells

Getting Started With Dumbbells

The use of dumbbells gives you a much more comprehensive strengthening effect because the workout engages your stabilizer muscles, in addition to the muscle you may be pin-pointing. Without all of the belts and artificial stabilizers of a machine, you also engage your core muscles, which are your body's natural stabilizers.

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