<p>Trying to develop a complete study guide--please show all work/steps in your

Question

<p>Trying to develop a complete study guide--please show all work/steps in your process so that I may easily work through it/follow. Thank you very much!<br /><br />A woman discus thrower exerts a constant torque on a discus of mass m = 1.00 kg over a total angular displacement<br />of +3&#960; radians (one and one-half body rotations in the counterclockwise direction). She thereby brings the discus to<br />a high velocity on its release. Assume that the discus is being rotated in a horizontal plane, parallel to the ground,<br />and that the distance of the discus from the center of rotation (the center of the woman's body) is 1.20 m. The discus<br />starts from rest.</p>
<p>(a) If the discus thrower generates a constant angular acceleration of +8.00 rad/s2 over the&#160;3&#960; radians of angular<br />distance before the release of the discus, what will be the angular velocity and the tangential speed of the discus<br />at the moment of release?</p>

Explanation / Answer

Radius of the rotation r = 1.20 m Total angular dispalcement = 3 rad Angular acceleration = 8.0 rad/s2 Initial angular speed i = 0 rad/s ---------------------------------------------------------------------------------- From rotaional kinematics                f2 = i2 + 2    Therefore the final angular velocity is                 f = 2                              = 2( 8.0 rad/s2 )( 3 rad)                       = 12.28 rad/s And the tangential speed is                 vt = r                    = (1.20 m) (12.28 rad/s)                    = 14.74 m/s                    = (1.20 m) (12.28 rad/s)                    = 14.74 m/s

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