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Pad 7:57 PM 75% a session.masteringphysics.com Women's Sweatshir... MyLab and Mastering Course Home MasteringPhysic... Course Home HW-9Practice Problem 9.3 previous | 5 of 15 l next Practice Problem 9.3 115.2, 15.2, Now let's consider the tangential and radial components of a discus as it is being thrown. A discus thrower turns with angular acceleration -50rad/52, moving the discus in a circle of radius 0.80 m . Find the radial and tangential components of acceleration of the discus (modeled as a point) and the magnitude of its acceleration at the instant when the angular velocity is 10 rad/s aran SOLUTION SET UP As shown in, we model the thrower's am as a rigid body, so r is constant (this may not be completely realistic). We model the discus as a particle moving in a circular path. Path of discus -50 rad/s s 10 rads SOLVE The components of the acceleration are given by u,2r = (1 0rad/s)" (0.80m) = 80 m/s2 ra-(0.80m)(50rad/s*) = 40 m/s2 r = 0.30 m and atan = O rad The magnitude of the acceleration vector is a-Varad2 + atan2 = 89 m/s Tan REFLECT The magnitude of the acceleration is about nine times the acceleration due to gravity, the corresponding force supplied by the thrower's arm must be about nine times the weight of the discus. Note that we have omitted the unit radian in our final results; we can do this because a radian is a dimensionless quantity. Part A - Practice Problem: what is the direction of the acceleration vector of the discus if the angular acceleration is 19 rad/s 2 and w = 12.0? Express your answer in degrees to two significant figures. Measure the angle from the radial line between the center of the discus and the thrower's shoulder Submit My Answers Give Up ContinueExplanation / Answer
Given,
alpha = 19 rad/s^2 ; w = 12
a(rad) = w^2r = 12^2 x 0.8 = 115.2 m/s^2
a(tan) = r alpha = 0.8 x 19 = 15.2 m/s^2
theta = tan^-1(15.2/115.2) = 7.52 deg deg
Hence, theta = 7.52 deg
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