A uniform circular disk (merry-go-round) of radius R=1.70 m andmass M=36.0 kg fr

Question

A uniform circular disk (merry-go-round) of radius R=1.70 m andmass M=36.0 kg freely rotates about a vertical axis, which isperpendicular to the ground, with initial angular velocityi=2.60 rad/s. A cat of mass m=5.90 kg, climbing ina tree, which hangs over the disk, falls straight down onto theedge of the disk and rides on it.

a) What is the magnitude of the angular velocity of the cat/disksystem after the cat has landed?

b) The cat decides to walk inward and sits down R/2 m from thecenter. What is now the magnitude of the angular velocity of thecat/disk system

Explanation / Answer

M= 36.0 R= 1.70 m m= 5.90 kg i=2.60 rad/s. a) What is the magnitude of the angularvelocity of the cat/disk system after the cat haslanded?
I= 0.5 MR2 (for a disk) Ic= m R2 (for a point mass) Angular momentum is conserved Ii =(I+Ic) then 0.5M R2 i =(0.5M R2 +mR2 ) = M i / (M +2m ) = 36 x 2.6 / (36 + 2 x 5.90) . = 1.96 rad/s
b) The cat decides to walk inward and sitsdown R/2 m from the center. What is now the magnitude of theangular velocity of the cat/disk system (I + Ic ) =(I+Ic')' where Ic' = m(R/2)2  then (0.5M R2 + m R2) =(0.5M R2 +m R2/4 )' (M + 2m ) = (M +m /2)' ' = (M + 2m )/ (M +m /2) ' = (36.0 + 2x 5.9)x 1.96 / ( 36.0+ 5.90/2 ) . ' =2.40 rad/s Please let me know if you have any questions. Please let me know if you have any questions.

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