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Unformatted text preview: CHAPTER 13 ROTATIONAL VECTORS AND ANGULAR MOMENTUM ActivPhysics can help with these problems: Activities 7.1, 7.16, 7.17 Section 131: Angular Velocity and Acceleration Vectors Problem 1. A car is headed north at 70 km/h. Give the magnitude and direction of the angular velocity of its 62cmdiameter wheels. Solution If we assume that the wheels are rolling without slipping (see Section 125), the magnitude of the angular velocity is = v cm /r = (70 m / 3 . 6 s) (0 . 31 m) = 62 . 7 s 1 . With the car going north, the axis of rotation of the wheels is eastwest. Since the top of a wheel is going in the same direction as the car, the righthand rule gives the direction of as west. Problem 2. If the car of Problem 1 makes a 90 left turn lasting 25 s, determine the average angular acceleration of the wheels. Solution If we assume that the turn is negotiated at a constant speed, then the magnitude of is also a constant 62 . 7 s 1 (see solution to Problem 1). However, the direction changes from west to south. Therefore av = / t has magnitude 2 / t = 2(62 . 7 s 1 ) / 25 s = 3 . 55 s 2 in a southeast direction. Problem 3. A wheel is spinning at 45 rpm with its spin axis vertical. After 15 s, its spinning at 60 rpm with its axis horizontal. Find (a) the magnitude of its average angular acceleration and (b) the angle the average angular acceleration vector makes with the horizontal. Solution Suppose that the x axis is horizontal in the direction of the final angular velocity ( f = (60 rpm) ) and the Problem 2 Solution. y axis is vertical in the direction of the initial angular velocity ( = (45 rpm) ) . Equation 131 implies that av = ( f i ) / t = (60 45 ) rpm / 15 s = (4 3 ) rpm / s . Its magnitude is av = radicalbig (4) 2 + ( 3) 2 rpm / s = 5 rpm / s = 5( / 30) s 2 = . 524 s 2 , at an angle = tan 1 ( 3 4 ) = 36 . 9 to the x axis (i.e., below the horizontal). Problem 4. A wheel is spinning about a horizontal axis, with angular speed 140 rad/s and with its angular velocity pointing east. Find the magnitude and direction of its angular velocity after an angular acceleration of 35 rad / s 2 , pointing 68 west of north, is applied for 5.0 s. Solution Take the x axis east and the y axis north, with positive angles measured CCW from the x axis. For constant angular acceleration, f = i + t = (140 s 1 ) + (35 s 2 )( cos158 + sin158 )(5 s) = ( 22 . 3 + 65 . 6 ) s 1 . Thus,  f  = 69 . 2 s 1 and f = 109 , or 18 . 8 west of north....
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This note was uploaded on 10/31/2010 for the course PHYSICS 69054 taught by Professor Huang during the Fall '09 term at University of California, Berkeley.
 Fall '09
 Huang

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