\"A horizontal platform in the shape of a circular diskrotates on a frictionless

Question

"A horizontal platform in the shape of a circular diskrotates on a frictionless bearing about a vertical axlethrough the center of the disk. The platform hasa radius of 1.10m and a rotational inertia of 486 kg * m^2about the axis of rotation. a 76.7 kg student walks slowlyfrom the rim of the platform toward the center. If theangular speed of the system is 2.34 rad/s when the studentstarts at the rim, what is the angular speed when she is .332m fromthe center?" Thank You!! "A horizontal platform in the shape of a circular diskrotates on a frictionless bearing about a vertical axlethrough the center of the disk. The platform hasa radius of 1.10m and a rotational inertia of 486 kg * m^2about the axis of rotation. a 76.7 kg student walks slowlyfrom the rim of the platform toward the center. If theangular speed of the system is 2.34 rad/s when the studentstarts at the rim, what is the angular speed when she is .332m fromthe center?" Thank You!!

Explanation / Answer

We know , angular momentum = angular velocity *moment ofinertia Initially , angular velocity = 2.34 rad/sec Also total moment of inertia =moment of inertia of disk + moment ofinertia of student                                                 = 486 + mr2 = 486 + 76.7*1.102 = 578.8kgm2 Finally , total moment of inertia =moment of inertia of disk + moment ofinertia of student                                                 = 486 + mr2 = 486 + 76.7*0.3322 =494.45 kgm2 From law of conservation of angular momentum , Initial angular momentum = final angular momentum 578.8*2.34 =494.45* final angular velocity final angular velocity =2.74 rad/sec

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