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

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11.1 ROTATIONAL VECTORS AND ANGULAR MOMENTUM EXERCISES 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 36 031 627 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 as west. 13. INTERPRET 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. DEVELOP 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 = v r cm m/s m rad/s 19 4 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 r i as west. In unit-vector notation, we write r ωω i i =− ˆ . After making a left turn, the angular speed remains unchanged, but the direction of r f is now south (see sketch). In unit-vector notation, we write r f j ˆ . EVALUATE Using Equation 10.4, we find the angular acceleration to be r r rr α av = −− ∆∆ tt ji i fi ˆ ( ˆ ) ( ˆˆ j ) The magnitude of r av is || (. ) . r av 2 rad/s s rad/s = 2 2627 25 355 t and r av points in the south-east direction (in the direction of the vector ). ij ASSESS Angular acceleration r av points in the same direction as r . 14. Suppose that the x -axis is horizontal in the direction of the final angular velocity (( ) ˆ ) f i = 60 rpm and the y -axis is vertical in the direction of the initial angular velocity ) ˆ ˆ i = 45 rpm j Equation 11.1 implies that αω av = () / t ( )( )/ . 60 45 15 4 3 i j rpm/ s rpm s = Its magnitude is av rpm s =+ = () ( ) / 43 22 5 5 30 0 524 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. INTERPRET The problem asks about the angular velocity of the wheels after an angular acceleration has been applied within a time interval. DEVELOP 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 r ω i and the angular acceleration r α can be expressed as r r ωω αα θ θ αα ii ij == =+ = ˆ () ˆ (cos ˆ sin ˆ ) 140 rad/s ( ˆˆ 35 rad/s )[cos(90 68 ) sin(90 68 ) ] 2 °+ ° + ° = ( . ˆ (. ˆ −+ 32 45 13 11 rad/s ) 22 The final angular velocity can be found by using Equation 10.8. EVALUATE Using Equation 10.8, the angular velocity at t = 50 .s is rr r ωωα fi ti i =+= + + ˆ [( . ˆ 140 32 45 rad/s rad/s ) 2 ˆ ]( . ) ˆ 13 11 5 0 22 3 65 rad/s ) s rad/s) 2 j i =− + 6rad/s) ˆ j The magnitude and direction of r f are ff + = || ( . . r 22 3 65 6 69 2 rad /s and f fy fx = = −− tan .
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This note was uploaded on 10/21/2010 for the course PHYSICS 2131441 taught by Professor Pheong during the Fall '10 term at Berkeley.

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

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