1. Chapter 7: Rotational Motion and the Law of Gravity

3. Section 7-1: Measuring Rotational Motion When an object spins, it is said to undergo rotational motion. Rotational motion– the motion of a body that spins about an internal axis. The axis of rotation is the line about which the rotation occurs.

4. A point on an object that rotates about a single axis undergoes circular motion. Circular motion– the motion of a body that rotates around an external axis. It is difficult to describe the motion of a point moving in a circle using only the linear quantities introduced in chapter 2 because the direction of motion in a circular path is constantly changing. For this reason, circular motion is described in terms of the angle through which the point on an object moves.

5. In the situations we have encountered so far, angles have been measured in degrees. In science, angles are often measured in radians (rad) rather than in degrees. Almost all of the equations in this chapter and the next require that angles be measured in radians. Radian– an angle whose arc length (s) is equal to its radius (r), which is approximately equal to 57.3°.

6. In general, any angle θ measured in radians is defined by the following: θ = s/r The radian is a pure number with no dimensions because θ is the ratio of an arc length (a distance) to the length of the radius (also a distance). When an object moves through an angle of 360° (one revolution), the arc length s is equal to the circumference of the circle 2πr. 1 revolution = 360° = 2π rad One revolution corresponds to an angle of approximately 6.28 rad (2 x 3.14).

7. Just as an angle in radians is the ratio of the arc length to the radius, the angular displacement traveled by the object is the change in the arc length, Δs, divided by the distance of the object from the axis of rotation. Angular displacement– the angle through which a point, line, or body is rotated in a specified direction and about a specified axis. Δθ = Δs/r For this book, when a rotating object is viewed from above, the arc length, s, is considered positive when the point rotates counterclockwise and negative when it rotates clockwise.

8. Angular speed– the rate at which a body rotates about an axis, usually expressed in radians per second. The average angular speed of a rotating rigid object is the ratio of the angular displacement, Δθ, to the time interval, Δt, that the object takes to undergo that displacement. Angular speed describes how quickly the rotation occurs.

9. ω avg = Δθ/Δt Angular speed is represented by the Greek letter omega. Angular speed is given in units of radians per second (rad/s). Sometimes angular speeds are given in revolutions per unit time. Recall that 1 rev = 2π rad.

10. Angular acceleration– the time rate of change of angular speed, expressed in radians per second squared (rad/s 2 ). Average angular acceleration is given by the following relationship: α avg = Δω/Δt Angular acceleration is the Greek letter alpha. Remember that all points on a rotating rigid object have the same angular acceleration and angular speed.

11. Nearly every linear quantity we have encountered thus far has a corresponding twin in rotational motion. The formulas for uniform or constant linear acceleration can also be applied for constant rotational acceleration. The terms for linear motion are replaced with those for angular motion.

12. Section 7-2: Tangential and Centripetal Acceleration It is useful to understand how the angular speed and angular acceleration of a rotating object relate to the linear speed and linear acceleration of a point on the object. Imagine a carousel rotating about its center. Because a carousel is a rigid object, any two horses attached to the carousel have the same angular speed and angular acceleration regardless of their respective distances from the axis of rotation.

13. However, if the two horses are different distances from the axis of rotation, they have different tangential speeds. Tangential speed– the instantaneous linear speed of an object directed along the tangent to the object’s circular path. The tangential speed of a horse on the carousel is its speed along a line drawn tangent to its circular path. Remember, the tangent to a circle is the line that touches the circle at one and only one point.

14. The horse on the outside must travel the same angular displacement during the same amount of time as the horse on the inside. To achieve this, the horse on the outside must travel a greater distance, Δs, than the horse on the inside. An object that is farther from the axis of a rigid rotating body must travel at a higher tangential speed around the circular path, Δs, to travel the same angular displacement as would an object closer to the axis.

15. The formula for determining tangential speed is: v t = rω This equation is derived from existing equations. The ω is the instantaneous angular speed, rather than the average angular speed, because the time interval is so short. This equation is valid only when the angular speed,ω, is measured in radians per second. Other measures of angular speed, such as degrees per second and revolutions per second, must not be used in this equation.

16. If a carousel speeds up, the horses on it experience an angular acceleration. The linear acceleration related to this angular acceleration is tangent to the circular path and is called the tangential acceleration. Tangential acceleration– the instantaneous linear acceleration of an object directed along the tangent to the object’s circular path. The formula is a t = rα This equation must use the unit radians to be valid. This equation is also derived from existing equations.

17. Centripetal Acceleration Remember that acceleration depends on a change in the velocity. Because velocity is a vector, there are two ways an acceleration can be produced: By a change in the magnitude of the velocity and By a change in the direction of the velocity.

18. For a car moving in a circular path with constant speed, the acceleration is due to a a change in direction. An acceleration of this nature is called a centripetal (center seeking) acceleration. Centripetal acceleration– acceleration directed toward the center of a circular path. The formula is: a c = v t 2 /r Because the tangential speed is related to the angular speed through the relationship v t = rω, the centripetal acceleration can be found using angular speed as well: a c = rω 2

19. Centripetal and tangential acceleration are not the same. The tangential component of acceleration is due to changing speed. The centripetal component of acceleration is due to changing direction. When both components of acceleration exist simultaneously, the tangential acceleration is tangent to the circular path and the centripetal acceleration points toward the center of the circular path. Because these components of acceleration are perpendicular to each other, the magnitude of the total acceleration can be found using the Pythagorean theorem: a total = (a t 2 + a c 2 ) ½

20. Find the total acceleration using the Pythagorean theorem: The direction of the total acceleration can be found using the inverse of the tangent function. θ = tan -1 (a c /a t )

21. Section 7-3: Causes of Circular Motion Consider a ball of mass m tied to a string of length r that is being whirled in a horizontal circular path. Assume that the ball moves with constant speed. Because the velocity vector, v, changes direction continuously during the motion, the ball experiences a centripetal acceleration directed toward the center of motion.

22. The inertia of the ball tends to maintain the ball’s motion in a straight line path; however, the string counteracts this tendency by exerting a force on the ball that makes the ball follow a circular path. This force is directed along the length of the string toward the center of the circle. The magnitude of this force can be found by applying Newton’s second law along the radial direction. The force producing centripetal acceleration is called centripetal force.

23. The formula for calculating centripetal force is F c = ma c Other forms of this formula can be obtained for use in calculations: F c = mv t 2 /r F c = mrω 2 The SI unit for centripetal force is the newton.

24. Picture a bike moving along the straightaway of an oval track. Because its speed is constant in a straight line, the bike is not accelerating. However, when the bike enters a curve, even if its speed does not change, it is accelerating because its direction is changing. In order for the bicycle to be accelerating, some unbalanced force must be acting on it in a direction toward the center of the curve. That force is a centripetal force.

25. No centripetal force means no circular motion. When a car rounds a sharp curve on a highway, the centripetal force is the friction between the tires and the road surface. If the road is icy or wet, and the tires, lose their grip, the centripetal force may not be enough to overcome the car’s inertia. Then the car would shoot off in a straight line in the direction it was traveling at the spot where it lost traction.

26. If the centripetal force is removed from an object moving in a circular path, the object will no longer move in its circular path. Initially, it will move along a straight-line path tangent to the circle, but then it will move in the parabolic path of a projectile.

27. To better understand the motion of a rotating system, consider a car approaching a curved exit ramp to the left at a high speed. As the driver makes the sharp left turn, the passenger slides to the right and hits the door. At that point, the force of the door keeps the passenger from being ejected from the car. What causes the passenger to move toward the door?

28. A popular explanation is that there must be a force that pushes the passenger outward. This force is sometimes called the centrifugal force, but that term often creates confusion, so it is not used in this textbook. Centrifugal force is a fictitious force. Centrifugal force– a so-called force that tends to move the particles of a spinning object away from the spin axis.

29. The phenomenon is correctly explained as follows: Before the car enters the ramp, the passenger is moving in a straight-line path. As the car enters the ramp and travels along a curved path, the passenger, because of inertia, tends to move along the original straight-line path. This is in accordance with Newton’s 1 st Law. However, if a sufficiently large force that maintains circular motion (toward the center of curvature) acts on the passenger, the person moves in a curved path, along with the car.

30. The origin of the force that maintains the circular motion of the passenger is the force of friction between the passenger and the car seat. If this frictional force is not sufficient, the passenger slides across the seat as the car turns underneath. Because of inertia, the passenger continues to move in a straight-line path. Eventually, the passenger encounters the door, which provides a large enough force to enable the passenger to follow the same curved path as the car.

31. The passenger slides toward the door not because of some mysterious outward force but because the force that maintains circular motion is not great enough to enable the passenger to travel along the circular path followed by the car. Inertia is often misinterpreted as centrifugal force.

32. Newton’s Law of Universal Gravitation The planets move in nearly circular orbits around the sun. The force that keeps these planets from coasting off in a straight line is a gravitational force. The gravitational force is a field force that always exists between two masses, regardless of the medium that separates them. Gravitational force– the mutual force of attraction between particles of matter.

33. It exists not just between large masses like the sun, Earth, and moon but between any two masses, regardless of size or composition. Gravitational force acts such that objects are always attracted to one another. Newton’s 3 rd law reminds us that these forces of attraction are always equal and opposite.

34. If masses m 1 and m 2 are separated by distance r, the magnitude of the gravitational force is given by the following equation: F g = G m 1 m 2 /r 2 G is a universal constant called the constant of universal gravitation. G = 6.673 x 10 -11 N·m 2 /kg 2 The law of universal gravitation is an example of an inverse-square law, because the force varies as the inverse square of the separation. The force between two masses decreases as the masses move farther apart.

35. The gravitational force exerted by a spherical mass on a particle outside the sphere is the same as it would be if the entire mass of the sphere were concentrated at its center. For example, the force on an object of mass m at Earth’s surface has the following magnitude: F g = G M E m/R E 2 M E is the mass of the earth– 5.98 x 10 24 kg. R E is the radius of the earth– 6.37 x 10 6 m This force is directed toward the center of earth. This force is in fact the weight of the mass, mg.

36. mg = G M E m/R E 2 By substituting the actual values for the mass and radius of Earth, we can find the value for g and compare it with the value of free-fall acceleration used throughout this book. This value for g is approximately equal to the value used throughout the book. The difference is due to rounding the values for Earth’s mass and radius.