<p>Trying to build a study guide--please show all steps so I may easily follow y

Question

<p>Trying to build a study guide--please show all steps so I may easily follow your process. Thank you!</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 />(c) What will be the magnitude of the total acceleration of the discus at the moment when 2&#960; radians of rotation are<br />complete? What will be the magnitude of the net force on the discus at this moment?</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 ----------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------- 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 -------------------------------------------------------------------------------------------------------------- Therefore  total acceleration of the discus at the moment when 2 radians of rotation are
complete is            a = ac2 +at 2                  = (120 m/s2) 2 + (9.6 m/s2)2                  = 120.38 m/s2   Therefore the force is            F = ma = ( 1.0 kg) (120.38 m/s2)                          = 120.38 N Therefore the force is            F = ma = ( 1.0 kg) (120.38 m/s2)                          = 120.38 N

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