A uniform solid disk, a uniform solid sphere, and a uniform hoop are placed side

Question

A uniform solid disk, a uniform solid sphere, and a uniform hoop are placed side by side at the top of an indline of height h. They are released from rest and roll without slipping, Place the objects in order of fastest to slowest at the bottom the incline. (Be sure to be able to explain why, in words, without equations.) uniform hoop, solid sphere, solid disk solid disk, solid sphere, uniform hoop the two spheres tie for first, followed by hoop solid disk, uniform hoop, solid sphere the hoop, followed by the two spheres in a tie it's a 3-way tie solid sphere, solid disk, uniform hoop uniform hoop, solid disk, solid sphere solid sphere, uniform hoop, solid disk Verify your answer by deriving a formula for their speeds when they reach the bottom in terms of h. (Use h and g as appropriate in your equations. For example: sqrt(2gh) means take the square root of 7*g#A) solid disk speed uniform hoop speed solid sphere speed Symbolic formating hep Now it's a race to the bottom among a golf ball, a bowling ball, and a marble (assume they are all uniform solid spheres, but of different masses, sizes, and densities). Place them in order from fastest to slowest at the bottom of the incline. golf ball, marble, bowling ball marble, bowling ball, golf ball golf ball, bowling ball, marble bowling ball, golf ball, marble marble, golf ball, bowling ball it's a 3-way tie bowling ball, marble, golf ball

Explanation / Answer

a)

I ( solid disk) =mr^2/ 2

I (solid sphere) =( 2/5mr^2)

I ( hoop)=mr^2

solid sphere,solid disk and in last hoop..the momnet of inertia is least for sphere , less the inertia , more the faster

we will use conservation of energy to derive the expression of velocity for three objects at the bottom of slide,

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

I ( solid disk) =mr^2/ 2

I (solid sphere) =( 2/5mr^2)

I ( hoop)=mr^2

for solid disk

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

gh = ( 1/4) ( v^2) + 1/2v^2

v^2= gh/ 0.75 or v = sqroot ( gh/0.75)

for solid sphere

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

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

v =sqroot ( gh/0.7)

for hoop

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

v = sqroot ( gh)

c) it's a three-way tie because they have the same inertia

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