A person with mass m p = 76 kg stands on a spinning platform disk with a radius

Question

A person with mass mp = 76 kg stands on a spinning platform disk with a radius of R = 1.89 m and mass md = 190 kg. The disk is initially spinning at = 1.6 rad/s. The person then walks 2/3 of the way toward the center of the disk (ending 0.63 m from the center).

4 ) What is the change in the total kinetic energy of the person and disk? (A positive value means the energy increased.)

5) What is the centripetal acceleration of the person when she is at R/3?

6) If the person now walks back to the rim of the disk, what is the final angular speed of the disk?

Explanation / Answer

initial postion of person = 3/2*(0.63) =0.945 m

conservation of angular momentum

(190)*1.89^2 /2 + 76*(0.945^2) ) *1.6 = (190)*1.89^2 /2 + 76*(0.63^2) ) w

=1.76326530612 rad/s

change in KE of person= 0.5 * (76*(0.63^2))(1.76326530612)^2 - 0.5 * 76*(0.945^2)(1.6^2)

=-39.9813355103 J

decreased

change in KE of disk = 0.5 * ((190)*1.89^2 /2) * (1.76326530612 )^2 - 0.5* ( ((190)*1.89^2 /2)) (1.6^2)

=93.1691755087 J

total change in KE of person + disk = 93.1691755087 -39.9813355103 =53.1878399984 J

means total KE has increased

5) v^2 / r = (wr)^2 / r = w^2 R = (1.76326530612 )^2 * 0.63 = 1.95873586005 m/s2

6)

conservation of angular momentum

(190)*1.89^2 /2 + 76*(0.945^2) ) *1.6 = (190)*1.89^2 /2 + 76*(1.89^2) ) w

=1.06666666667 rad/s

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