chap 11 - ROTATIONAL VECTORS AND ANGULAR MOMENTUM 11...

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11.1 ROTATIONAL VECTORS AND ANGULAR MOMENTUM E XERCISES Section 11.1 Angular Velocity and Acceleration Vectors 12. If we assume that the wheels are rolling without slipping (see Section 10.5), the magnitude of the angular velocity is ω = = = v cm m/ s m s / ( . )/( . ) . . r 70 3 6 0 31 62 7 1 With the car going north, the axis of rotation of the wheels is east- west. Since the top of a wheel is going in the same direction as the car, the right-hand rule gives the direction of ω as west. 13. I NTERPRET The problem asks about the angular acceleration of the wheels as the car traveling north with a speed of 70 km/h makes a 90 ° left turn that lasts for 25 s. D EVELOP The speed of the car is v cm km/h m/s. = = 70 19 4 . Assuming that the wheels are rolling without slipping, the magnitude of the initial angular velocity is ω = = = v r cm m/s m rad/s 19 4 0 31 62 7 . . . With the car going north, the axis of rotation of the wheels is east-west. Since the top of a wheel is going in the same direction as the car, the right-hand rule gives the direction of ω i as west. In unit-vector notation, we write ω ω i i = − ˆ . After making a left turn, the angular speed remains unchanged, but the direction of ω f is now south (see sketch). In unit-vector notation, we write ω ω f j = − ˆ . E VALUATE Using Equation 10.4, we find the angular acceleration to be α ω ω ω ω ω ω av = = = − − = t t j i t t i f i ˆ ( ˆ ) ( ˆ ˆ j ) The magnitude of α av is | | ( . ) . α ω av 2 rad/s s rad/s = = = 2 2 62 7 25 3 55 t and α av points in the south-east direction (in the direction of the vector ˆ ˆ ). i j A SSESS Angular acceleration α av points in the same direction as ω . 14. Suppose that the x -axis is horizontal in the direction of the final angular velocity ( ( ) ˆ ) ω f i = 60 rpm and the y -axis is vertical in the direction of the initial angular velocity ( ( ) ˆ ). ˆ ω i = 45 rpm j Equation 11.1 implies that α ω ω av = = ( )/ f i t ( ˆ ˆ ) ( ˆ ˆ ) / . 60 45 15 4 3 i j i j rpm/ s rpm s = Its magnitude is α av rpm s = + − = ( ) ( ) / 4 3 2 2 5 5 30 0 524 2 2 rpm s s s / ( / ) . , = = π at an angle θ = = − ° tan ( ) . 1 3 4 36 9 to the x axis (i.e., below the horizontal). 11
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11.2 Chapter 11 15. I NTERPRET The problem asks about the angular velocity of the wheels after an angular acceleration has been applied within a time interval. D EVELOP Take the x -axis east and the y -axis north, with positive angles measured CCW from the x -axis. In unit- vector notation, the initial angular velocity ω i and the angular acceleration α can be expressed as ω ω α α θ θ α α i i i i i j = = = + = ˆ ( ) ˆ (cos ˆ sin ˆ ) 140 rad/s ( ˆ ˆ 35 rad/s )[cos(90 68 ) sin(90 68 ) ] 2 ° + ° + ° + ° = i j ( . ˆ ( . ˆ + 32 45 13 11 rad/s ) rad/s ) 2 2 i j The final angular velocity can be found by using Equation 10.8. E VALUATE Using Equation 10.8, the angular velocity at t = 5 0 . s is ω ω α f i t i i = + = + + ( ) ˆ [( .
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