<p>Trying to build a solid study guide-please show all work/steps so that I may

Question

<p>Trying to build a solid study guide-please show all work/steps so that I may work through your process easily. Thank you so much!</p>
<p>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.<br /><br />(b) What will be the centripetal and tangential&#160;accelerations of the discus at the moment when 2&#960; radians of rotation are complete?</p>

Explanation / Answer

Radius of the rotation r = 1.20 m Total angular dispalcement = 2 rad Angular acceleration = 8.0 rad/s2 Initial angular speed i = 0 rad/s ---------------------------------------------------------------------------------- From rotaional kinematics                f2 = i2 + 2    Thenthe final angular velocity after = 2 rad is                 f = 2                              = 2( 8.0 rad/s2 )( 2 rad)                       = 10.0 rad/s And the tangential speed is                 vt = r                    = (1.20 m) (10 rad/s)                    = 12 m/s ----------------------------------------------------------------------------------------------- Therefore the centripetal acceleration after 2 rad                 ac = v2/r                     = ( 12 m/s)2 / 1.2 m                     = 120 m/s2 And tangential acceleration after 2 rad                 at = r                     = (1.2 m) ( 8.0 rad/s2 )                     = 9.6 m/s2 Total angular dispalcement = 2 rad Angular acceleration = 8.0 rad/s2 Initial angular speed i = 0 rad/s ---------------------------------------------------------------------------------- From rotaional kinematics                f2 = i2 + 2    Thenthe final angular velocity after = 2 rad is                 f = 2                              = 2( 8.0 rad/s2 )( 2 rad)                       = 10.0 rad/s And the tangential speed is                 vt = r                    = (1.20 m) (10 rad/s)                    = 12 m/s ----------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------- Therefore the centripetal acceleration after 2 rad                 ac = v2/r                     = ( 12 m/s)2 / 1.2 m                     = 120 m/s2 And tangential acceleration after 2 rad                 at = r                     = (1.2 m) ( 8.0 rad/s2 )                     = 9.6 m/s2                     = ( 12 m/s)2 / 1.2 m                     = 120 m/s2 And tangential acceleration after 2 rad                 at = r                     = (1.2 m) ( 8.0 rad/s2 )                     = 9.6 m/s2

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