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eargulansp 10. A computer is reading data ro the disc,the centripetal accelerati

ID: 1773724 • Letter: E

Question

eargulansp 10. A computer is reading data ro the disc,the centripetal acceleration 120 0.050 m from the center of the disc? om a rotating CD-ROM. At a point that iso.030 m/from the center of n is 120 m/s What is the centripetal acceleration at a point that is 2. Le 0.030 2 1 14. At an amusement park there is a ride in which cylindrically shaped chambers spin around a central axis. People sit in seats facing the axis, their backs against the outer wall. At one instant the outer wall moves at a speed of3.2 m/s, and an 83-kg person feels a 560-N force pressing against his back. What is the radius of the chamber?

Explanation / Answer

formula for centripetal accleration is

a= v^2/r

a1 = v^2/r1

a2 = v^2/ r2

a2 = a1 ( r2/ r1) = 120( 0.05/0.03) =200 m/s^2

(2)

formula for centripetal force is

Fc= mv^2/ r

r = mv^2/ Fc

=83 kg ( 3.2)^2/560 N

=1.5 m