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Unformatted text preview: A disc of radius 50.0 cm is rotating at 120 rpm, and its angular acceleration is 3.00 rad/s2. Find the magnitudes of the velocity, v, and tangential acceleration, aT, of its outer edge. 1. 2. 3. 4. 5.
10/9/09 v = 3.14 m/s v = 3.14 m/s v = 60.0 m/s v = 188 m/s None of the above. aT = 0.75 m/s2 aT = 1.50 m/s2 aT = 3.00 m/s2 aT = 1.50 m/s2 Oregon State University PH 201, Lecture #6 1 Angular Position vs. Translational s=r Displacement s=r Velocity avg. = /t vavg. = s/ t = r /t=r /t=r
2 avg Acceleration /t avg. =
10/9/09 aT.avg. = v/ t = r
Oregon State University PH 201, Lecture #6 avg. Two runners are racing around a circular track in different lanes (1 and 8). Lane 1 is 60.0 m from the center of the circle. Lane 8 is 68.0 m from that center. If the runners each complete 4 laps (each at a steady speed) in the same time, 3:56, how much faster is runner 8’s pace than runner 1’s? 1. 2. 3. 4. 5.
10/9/09 0.852 m/s 0.891 m/s 1.17 m/s 8.99 m/s None of the above.
Oregon State University PH 201, Lecture #6 3 Same two runners in the same two lanes on the same track as previously—but it’s a new race: This time they start from rest and run just one lap, again ﬁnishing in a tie, in 49.3 seconds. If runner 1’s speed at the ﬁnish was 15.3 m/s, ﬁnd runner 8’s average angular acceleration ( ) for this race. 1. 2. 3. 4. 5.
10/9/09 2.09 x 103 rad/s2 2.37 x 103 rad/s2 1.03 x 101 rad/s2 1.17 x 101 rad/s2 None of the above.
Oregon State University PH 201, Lecture #6 4 Rotational (Angular) Kinematics = (1/2)( = f= 2= f t+ ( t)2 i i + ( t) 2+2 ( ) i
i+ (1/2) f) t (Circular Translational Kinematics) s = r( ) vi = r i vf = r f t= t
Oregon State University PH 201, Lecture #6 aT = r 10/9/09 5 The equations of kinematics for angular motion can be applied to the motion of a rotating body only when: • the angular acceleration, , is constant over the entire motion being analyzed (the whole interval t). • the signs of the vectors , i, f, and are observed and used properly! 10/9/09 Oregon State University PH 201, Lecture #6 6 A ladybug is standing on the surface of a disk at a distance of 20.0 cm from its center. Initially the disk is rotating clockwise so that the bug’s speed is 2.00 m/s. Then the disk undergoes a constant positive angular acceleration for 5.00 seconds. The bug’s speed is then 8.00 m/s. Find: The disk’s angular acceleration during the 5 seconds; the disk’s angular displacement during the 5 seconds. 1. 2. 3. 4. = –6.00 rad/s2 = –0.600 rad/s2 = 1.00 rad/s2 = 10.0 rad/s2 = –125 rad = –12.5 rad = 7.50 rad = 75.0 rad 5. None of the above.
10/9/09 Oregon State University PH 201, Lecture #6 7 ...
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 Fall '08
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 Physics, Acceleration

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