Holt Physics Chapter 7. Rotational Motion

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1 Holt Physics Chapter 7 Rotational Motion

2 Measuring Rotational Motion Spinning objects have rotational motion Axis of rotation is the line about which rotation occurs A point that moves around an axis undergoes circular motion (revolution)

3 s=arc length, r=radius =angle of rotation (measured in degrees or radians) corresponding to s s r

4 (rad) = s/r If s = r, then = 1 radian If s = 2 r (circumference), then (rad) = 2 r/r= 2 radians Therefore 360 degrees = 2 radians, or 6.28 See figure 7-3, page 245

5 s r

6 Equation to convert between radians and degrees (rad) = [ /180º] X (deg) Sign convention: clockwise is negative, counterclockwise is positive!

7 Angular Displacement ( ) describes how much an object has rotated = s/r

8 s r

9 Example 1. A child riding on a merry-go-round sits at a distance of m from the center. Calculate his angular displacement in radians, as he travels in a clock-wise direction through an arc length of 1.41 m.

10 Given: r = m Find: Solution: Δθ s r s = m -1.41m rad m The negative sign in our answer indicates direction.

11 Book Example pg. 246: While riding on a carousel that is rotating clockwise, a child travels through an arc length of 11.5m. If the child s angular displacement is 165 degrees, what is the radius of the carousel? = -165 degrees but we need radians! rad = [ /180º] X deg = ( /180º)(-165º) = radians s = m r=? = s/r so, r= s / r = -11.5/-2.88 = 3.99m

12 Angular Speed Angular Speed ( ave ) is equal to the change in angular displacement(in radians!) per unit time. ave = / t, in radians/sec Remember, One revolution = radians,

13 Therefore, if you want to know the number of revolutions: # of Revolutions = rad /2 rad Or rad = # of rev. X 2 rad

14 s r

15 Book Example pg. 248: A child at an ice cream parlor spins on a stool. The child turns counterclockwise with an average angular speed of 4.0 rad/s. In what time interval will the child s feet have an angular displacement of 8.0 radians. =8 rad ave = 4.0 rad/s t=? ave = / t, so t= / ave t= 8 rad / 4.0 rad/s = 2.0 s = 6.3 sec

16 Example 4. A man ties a ball to the end of a string and swings it around his head. The ball s average angular speed is 4.50 rad/s. A) In what time interval will the ball move through an angular displacement of 22.7 rad? B) How many complete revolutions does this represent?

17 Given: avg = 4.50 rad/s = 22.7 rad Find: A) t B) # of revolutions Solution: Δt # of rev. 2 θ ω avg θ rad 22.7 rad 5.04 s 4.50 rad/s 22.7 rad 2 rad 3.61rev.

18 Homework 7A page 247 and Practice 7B page 248

19 Angular Acceleration Angular Acceleration ( ave ) is the change in angular speed ( ) per unit time(s). ave = / t = ( 2-1 )/(t 2 -t 1 ) Unit is rad/s 2

20 Book Example 1: A car s tire rotates at an initial angular speed of 21.5 rad/s. The driver accelerates, and after 3.5 sec the tire s angular speed is 28.0 rad/s. What is the tire s average angular acceleration during the 3.5 s time interval? 1 =21.5 rad/s 2 =28.0 rad/s t=3.5 s ave =? ave = 2-1 / t = ave =(28.0rad/s-21.5rad/s)/3.5s ave =1.9 rad/s 2

21 Example 2. The angular velocity of a rotating tire increases from 0.96 rev/s to rev/s with an average angular acceleration of 6.0 rad/s 2. Find the time required for given angular acceleration. First change revolutions/sec to radians/sec by multiplying by 2 rad. (0.96rev/s)X 2 rad = 6.0 rad/sec (1.434rev/s)X 2 rad = 9.01 rad/sec

22 Given: 1 = 6.0 rad/s 2 =9.01 rad/s avg =6.0rad/s 2 Find: t =? Solution: avg ω2 1 - ω t t ω - ω avg t = 9.01 rad/s 6.0 rad/s = 6.0 rad/s s 2 1

23 Homework 7C pg 250

24 Angular Kinematics All points on a rotating rigid object have the same angular acceleration and angular speed. Angular and linear quantities correspond if angular speed ( ) is instantaneous and angular acceleration ( ) is constant. See table 7-2, pg. 251 or your formula sheet

25 Equations for Motion with Constant Acceleration Linear f i a t 1 x i t a t f 2a x i 2 Rotational ωf ωi α t 1 θ ωi t α t 2 ω f 2 ω i 2 2 θ 2 Δx 1 (ν 2 i νf)δt Δθ 1 (ω ω) f 2 i Δt

26 Book Example 1 (page 251)The wheel on an upside-down bicycle moves through 11.0 rad in 2.0 sec. What is the wheel s angular acceleration if its initial angular speed is 2.0 rad/s? =11.0 rad t=2.0 s i = 2.00rad/s =? Use one of the angular kinematic equations from Table 7-2 to solve for which one? i t ½ ( t) 2

27 Solve for =2( - i t)/( t) 2 =2[11.0rad-(2.00rad/s)(2.0s)]/(2.0s) 2 = 3.5 rad/s 2

28 Example 2. A CD initially at rest begins to spin, accelerating with a constant angular acceleration about its axis through its center, achieving an angular velocity of 3.98 rev/s when its angular displacement is 15.0 rad. what is the value of the CD s angular acceleration?

29 Given: i = 0.0 rad/s f = 3.98 rev/s X 2 = 25.0 rad/s = 15.0 rad Find: Original Formula: ω f 2 ω 2 i 2 θ The i = 0.0 rad/s, so

30 Given: i = 0.0 rad/s f = 25.0 rad/s = 15.0 rad Find: Formula: ω 2 f 2 θ Now, we isolate

31 Given: i = 0.0 rad/s f = 25.0 rad/s = 15.0 rad Find: Working Formula: α ω f 2 2 θ

32 Given: i = 0.0 rad/s f = 25.0 rad/s = 15.0 rad Find: Solution: α 2 ωf 2 θ (25.0 rad/s) 2(15.0 rad) rad/s 2

33 Homework Problems 7D, page 252

34 Review and Test

35 Tangential Speed Although any point on a fixed object has the same angular speed, their tangential speeds vary based upon their distance from the center or axis of rotation. Tangential speed (v t ) instantaneous linear speed of an object directed along the tangent to the objects circular path. (see fig. 7-6, pg. 253)

36 Formula v t = r Make sure is in radians/s r is in meters v t is in m/s

37 Book Example 1 (pg. 254): The radius of a CD in a computer is m. If a microbe riding on the disc s rim had a tangential speed of 1.88m/s, what is the disc s angular speed? r=0.0600m =? v t =1.88m/s Use the tangential speed equation (v t = r )to solve for angular speed. =v t /r =1.88m/s/0.0600m =31.3 rad/s

38 Example 2. A CD with a radius of 20.0 cm rotates at a constant angular speed of 1.91 rev/s. Find the tangential speed of a point on the rim of the disk. Given: Find: t r =20.0cm= m = 2 X1.91rev/s= 12.0 rad/s Solution: ν t rω (0.200 m)(12.0 rad/s) 2.40 m/s

39 Tangential Acceleration Tangential Acceleration (a t )- instantaneous linear acceleration of an object directed along the tangent to the objects circular path. Formula a t = r Make sure that a t is in m/s 2 r is in meters is in radians/s 2

40 Book Example pg.256: A spinning ride at a carnival has an angular acceleration of 0.50rad/s 2. How far from the center is a rider who has a tangential acceleration of 3.3 m/s 2? =0.50rad/s 2 a t = 3.3 m/s 2 r=? Use the tangential acceleration equation (a t = r ) and rearrange to solve for r. r=a t / r = (3.3 m/s 2 )/(0.50rad/s 2 ) r=6.6m

41 Example 2. The angular velocity of a rotating CD with a radius 20.0 cm increases from 2.0 rad/s to 6.0 rad/s in 0.50 s. What Is the tangential acceleration of a point on the rim of the disk during this time interval? Hint: Start by finding the angular acceleration

42 Given: r = m i = 2.0 rad/s f = 6.0 rad/s t = 0.50 s Find: ωf - ωi 6.0 rad/s rad/s 2 Solution: 8.0 rad/s t 0.50 s

43 Given: r = m = 8.0 rad/s 2 t = 0.50 s Find: Solution: a t 2 2 a t rα (0.200 m)(8.0 rad/s ) 1.6 m/s

44 7E, page 255, 7F page 256

45 Centripetal Acceleration Centripetal acceleration is the acceleration directed toward the center of a circular path. Caused by the change in direction Formula a c = v t2 /r And also a c = r 2

46 Example 1 pg. 258: A test car moves at a constant speed around a circular track. If the car is 48.2 m from the track s center and has a centripetal acceleration of 8.05 m/s 2. What is its tangential speed? r=48.2m a c = 8.05 m/s 2 v t =? Use the first centripetal acceleration equation (a c = v t2 /r) to solve for v t. v t2 = a c Xr v t = a c Xr v t = (8.05m/s 2 X48.2m) v t = 19.7m/s

47 Example 2. An object with a mass of kg, Moving in a circular path with a radius of 0.25 m, Experiences a centripetal acceleration of 8.0 m/s 2. Find the object s angular speed.

48 Example 2. An object with a mass of kg, Moving in a circular path with a radius of 0.25 m, Experiences a centripetal acceleration of 8.0 m/s 2. Find the object s angular speed. Given: m = kg r = 0.25 m a c = 8.0 m/s 2 Find: Original Formula: a c rω 2

49 Example 2. An object with a mass of kg, Moving in a circular path with a radius of 0.25 m, Experiences a centripetal acceleration of 8.0 m/s 2. Find the object s angular speed. Given: m = kg r = 0.25 m a c = 8.0 m/s 2 Find: Solution: ω a r c 8.0 m/s 0.25 m 2 32 /s rad/s

50 Homework: 7G pg. 258

51 Lock down drill!!!

52 Using the Pythagorean Theorem Tangential and centripetal acceleration are perpendicular, so you can use Pythagorean theorem and trig to find total acceleration and direction, if asked for. See figure 7-9, on page 259

53 a t a c a total

54 Suppose a ball (mass m) moves with a constant speed V while attached to a string.

55 Is there an outward force, pulling the ball out? NO! a t (and v t ) a T c

56 What about the centrifugal force I feel when my car goes around a curve at high speed? There is no such thing* as centrifugal force! You feel the centripetal force of the door pushing you towards the center of the circle of the turn! Your confused brain interprets the effect of Newton s first law (inertia) as a force pushing you outward. *Engineers talk about centrifugal force when they attach a coordinate system to an accelerated object. In order to make Newton s Laws work, they have to invent a pseudoforce, which they call centrifugal force.

57 Centripetal Force Centripetal force is necessary for circular motion. Without it, the object would go in a linear path tangent to the circular motion at that point. Then it would follow the motion of a free-falling body (parabolic path toward the ground).

58 Causes of Circular Motion Centripetal acceleration (a c ) is caused by a FORCE directed toward the center of the circular path. This force is called centripetal force (F c ) Formula F c =ma c F c =mv t2 /r or, F c =mr 2 The unit is the Newton (kg m/s 2 )

59 Book Example Pg. 261: A pilot is flying a small plane at 30.0m/s in a circular path with a radius of 100.0m. If a force of 635N is needed to maintain the pilot s circular motion, what is the pilot s mass? r=100.0m F c = 635 N v t =30.0m/s m=? Formula F c =mv t2 /r solve for m m=f c (r/v t2 ) m=635nx[100.0m/(30.0m/s) 2 ] m=70.6kg

60 Example 2: A raccoon sits 2.2 m from the center of a merry-go-round with an angular speed of 2.9 rad/s. If the magnitude of the force that maintains the raccoon s circular motion if 67.0 N, what is the raccoon s mass? r=2.2m F c = 67.0 N =2.9 rad/s m=? Formula F c =m r 2 solve for m m=f c /r 2 m=67.0n/[(2.2m)(2.9rad/s) 2 ] m=3.6kg

61 Newton s Law and Gravity Gravitational force depends on the mass of the objects involved. Formula F g = Gm 1 m 2 /r 2 G is the constant of universal gravitation = 6.673X10-11 Nm 2 /kg 2 Gravitational force is localized to the center of a spherical mass

62 Book Example pg. 264: Find the distance between a 0.300kg billiard ball and a 0.400kg billiard ball if the magnitude of the gravitational force is 8.92X10-11 N. m 1 =0.300kg m 2 =0.400kg r=? F g =8.92X10-11 N F g = Gm 1 m 2 /r 2 G=6.673X10-11 Nm 2 /kg 2 r 2 =(G/F g )m 1 m 2 r= [(6.673X10-11 Nm 2 /kg 2 )/8.92X10-11 N](0.30kgX0.400kg) r= 3.00X10-1 m

63 This is where we get our value for Free Fall Acceleration For the earth, F g = mg, and F g = mg= Gm E m/r E 2 So, g= Gm E /r E 2 If the earth s radius is 6.37X10 6 m, and the earth s mass is 5.98X10 24 kg, what is acceleration due to gravity? 9.83m/s 2 close!!

64 Homework 7H pg. 261, 7I pg. 265 and complete Amusement Park Physics Lab

65 Amusement Park Physics

66 Review Problems for Ch. 7 1,3,6,7,9,10,12,16,20,21,24,25, 30,38,40,43

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