1 Rotational Kinematics Rotational Motion and Angular Displacement Chapter 8 Lesson 3

2

3 Rolling Motion

4 Linear speed Linear acceleration Tangential speed, v T Tangential acceleration, a T

5 Example 8. An Accelerating Car An automobile starts from rest and for 20.0 s has a constant linear acceleration of m/s 2 to the right. During this period, the tires do not slip. The radius of the tires is m. At the end of the 20.0-s interval, what is the angle through which each wheel has rotated?

6 w w0 w0 t ? –2.42 rad/s 2 0 rad/s 20.0 s

7 The Vector Nature of Angular Variables Right-Hand Rule Grasp the axis of rotation with your right hand, so that your fingers circle the axis in the same sense as the rotation. Your extended thumb points along the axis in the direction of the angular velocity vector. Angular acceleration arises when the angular velocity changes, and the acceleration vector also points along the axis of rotation. The acceleration vector has the same direction as the change in the angular velocity.

8 Concepts & Calculations Example 9. Riding a Mountain Bike A rider on a mountain bike is traveling to the left. Each wheel has an angular velocity of rad/s, where, as usual, the plus sign indicates that the wheel is rotating in the counterclockwise direction.

9 (a)To pass another cyclist, the rider pumps harder, and the angular velocity of the wheels increases from to rad/s in a time of 3.50 s. (b)After passing the cyclist, the rider begins to coast, and the angular velocity of the wheels decreases from to rad/s in a time of 10.7 s. In both instances, determine the magnitude and direction of the angular acceleration (assumed constant) of the wheels. (a) The angular acceleration is positive (counterclockwise).

10 (b) The angular acceleration is negative (clockwise).

11 Concepts & Calculations Example 10. A Circular Roadway and the Acceleration of Your Car

12 Suppose you are driving a car in a counterclockwise direction on a circular road whose radius is r = 390 m (see Figure 8.20). You look at the speedometer and it reads a steady 32 m/s (about 72 mi/h). (a) What is the angular speed of the car? (b) Determine the acceleration (magnitude and direction) of the car. (c) To avoid a rear-end collision with a vehicle ahead, you apply the brakes and reduce your angular speed to 4.9 × 10 –2 rad/s in a time of 4.0 s. What is the tangential acceleration (magnitude and direction) of the car? (a)

13 (b) (c)

14 Problem 5 REASONING AND SOLUTION Using Equation 8.4 and the appropriate conversion factors, the average angular acceleration of the CD in rad/s 2 is The magnitude of the average angular acceleration is 6.4 × rad/s 2

15 Problem 7 REASONING AND SOLUTION Equation 8.4 gives the desired result. Assuming t 0 = 0 s, the final angular velocity is

16 Problem 13 REASONING AND SOLUTION The baton will make four revolutions in a time t given by Half of this time is required for the baton to reach its highest point. The magnitude of the initial vertical velocity of the baton is then

17 With this initial velocity the baton can reach a height of

18 Problem 14 REASONING AND SOLUTION The figure above shows the relevant angles and dimensions for either one of the celestial bodies under consideration.

19 a. Using the figure above -3

20 b. Since the sun subtends a slightly larger angle than the moon, as measured by a person standing on the earth, the sun cannot be completely blocked by the moon. Therefore,. c. The relevant geometry is shown below.

21 The apparent circular area of the sun as measured by a person standing on the earth is given by:, where R sun is the radius of the sun. The apparent circular area of the sun that is blocked by the moon is, where R b is shown in the figure above. Also from the figure above, it follows that R sun = (1/2) s sun and R b = (1/2) s b Therefore, the fraction of the apparent circular area of the sun that is blocked by the moon is The moon blocks out 95.1 percent of the apparent circular area of the sun.

22 Problem 17 REASONING AND SOLUTION Since the angular speed of the fan decreases, the sign of the angular acceleration must be opposite to the sign for the angular velocity. Taking the angular velocity to be positive, the angular acceleration, therefore, must be a negative quantity. Using Equation 8.4 we obtain

23 Problem 21 REASONING Equation 8.8 from the equations of rotational kinematics can be employed to find the final angular velocity . The initial angular velocity is  0 = 0 rad/s since the top is initially at rest, and the angular acceleration is given as  = 12 rad/s 2. The angle θ (in radians) through which the pulley rotates is not given, but it can be obtained from Equation 8.1 (θ = s/r ), where the arc length s is the 64-cm length of the string and r is the 2.0-cm radius of the top. SOLUTION Solving Equation 8.8 for the final angular velocity gives

24 We choose the positive root, because the angular acceleration is given as positive and the top is at rest initially. Substituting θ = s/r from Equation 8.1 gives

25 Problem 29 REASONING AND SOLUTION Equation 8.9 gives the desired result -3

26 Problem 39 REASONING Since the car is traveling with a constant speed, its tangential acceleration must be zero. The radial or centripetal acceleration of the car can be found from Equation 5.2. Since the tangential acceleration is zero, the total acceleration of the car is equal to its radial acceleration. SOLUTION a. Using Equation 5.2, we find that the car’s radial acceleration, and therefore its total acceleration, is

27 b The direction of the car’s total acceleration is the same as the direction of its radial acceleration. That is, the direction is

28 Problem 42 REASONING The drawing shows a top view of the race car as it travels around the circular turn. Its acceleration a has two perpendicular components: a centripetal acceleration a c that arises because the car is moving on a circular path and a tangential acceleration a T due to the fact that the car has an angular acceleration and its angular velocity is increasing.

29 We can determine the magnitude of the centripetal acceleration from Equation 8.11 as a c = r  2, since both r and  are given in the statement of the problem. As the drawing shows, we can use trigonometry to determine the magnitude a of the total acceleration, since the angle (35.0  ) between a and a c is given. SOLUTION Since the vectors ac and a are one side and the hypotenuse of a right triangle, we have that

30 The magnitude of the centripetal acceleration is given by Equation 8.11 as a c = r  2, so the magnitude of the total acceleration is

31 Problem 46 REASONING AND SOLUTION a. If the wheel does not slip, a point on the rim rotates about the axle with a speed v T = v = 15.0 m/s For a point on the rim  = v T /r = (15.0 m/s)/(0.330 m) b. v T = r  = (0.175 m)(45.5 rad/s) = =

32 Problem 50 REASONING The angle through which the tire rotates is equal to its average angular velocity multiplied by the elapsed time t,  t. According to Equation 8.6, this angle is related to the initial and final angular velocities of the tire by The tire is assumed to roll at a constant angular velocity, so that  0 =  and  =  t. Since the tire is rolling, its angular speed is related to its linear speed v by Equation 8.12, v = r , where r is the radius of the tire. The angle of rotation then becomes

33 The time t that it takes for the tire to travel a distance x is equal to t = x/v, according to Equation 2.1. Thus, the angle that the tire rotates through is SOLUTION Since 1 rev = 2  rad, the angle (in revolutions) is

34 Problem 51

35 REASONING Assuming that the belt does not slip on the platter or the shaft pulley, the tangential speed of points on the platter and shaft pulley must be equal; therefore, SOLUTION Solving the above expression for gives

36 Problem 60 REASONING AND SOLUTION a. The tangential acceleration of the train is given by Equation 8.10 as The centripetal acceleration of the train is given by Equation 8.11 as

37 The magnitude of the total acceleration is found from the Pythagorean theorem to be b. The total acceleration vector makes an angle relative to the radial acceleration of

38 Problem 67 REASONING AND SOLUTION By inspection, the distance traveled by the "axle" or the center of the moving quarter is

39 where r is the radius of the quarter. The distance d traveled by the "axle" of the moving quarter must be equal to the circular arc length s along the outer edge of the quarter. This arc length is, where is the angle through which the quarter rotates. Thus, so that. This is equivalent to