Problem 10.73 Part A Calculate its translational speed when it reaches the botto

Question

Problem 10.73 Part A Calculate its translational speed when it reaches the bottom. A sphere of radius ro = 23.0 cm and mass m = 1.20kg starts from rest and rolls without slipping down a 40.0° incline that is 12.0 m long. m/s u= Submit My Answers Give Up Part B Calculate its rotational speed when it reaches the bottom rad/s Submit My Answers Give Up Part C What is the ratio of translational to rotational kinetic energy at the bottom Submit My Answers Give Up Part D Doe s your answer in part A depend O depends on mass O depends on radius O doesnt depend on mass and radius O depends on mass and radius on mass or radius of the ball? Submit My Answers Give Up

Explanation / Answer

a height of incline= 12 sin 40=7.713 m apprx

at the bottom,

mgh = 1/2 mv^2 + 1/2Iw^2

I of sphere= 2/5 mr^2

mgh = ( 1/2)( 2/5 mr^2) (v^2/r^2) + 1/2 mv^2

gh = ( 1/5 v^2)+ ( 1/2v^2)

gh = ( 7/10) v^2

v = sqroot ( 10gh/7) = sqroot ( 10 x 9,8 x 7.713 / 7) = 10.39 m/s

b) w= v/r= 10.39/23 x 10^-2 = 45.18 rad/s

c)K transklional/ K rotatioal = ( 1/2 mv^2) / ( 1/2 Iw^2) = ( mv^2) / ( I w^2) = 5/2

d)doesn;t depend on mass or radius

e) radius

f) doesn't depend on mass or radius

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