Question

A uniform solid disk, a uniform solid sphere, and a uniform hoop are placed side by side at the top of an incline 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.) 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(7gh) means take the square root of 7*g*h) 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.

Answer 1

Sphere, disk, hoop

we assume all the objects have the same mass and radious. the MI of the sphere is the least and that of the hoop is the largest, they are more inert in that order, the least inert will have more speed or easy to change its state of rest or of motion.

golf ball, bowling ball and marble.

It is 3-way tie as the MI depends on the actual mass and the size or the radious of the sphere.

the least MI object will be the fastest.

speeds :

MI if sphere Is =2MR^2/5

Total KE (roatational + transitional)

KE = Is$\omega$²/2 +MV²/2 = (2MR^2/5)*(V/R)^2 /2 + MV^2/2 ( $\omega$ = V/R)

       =7MV^2/10

       = Mgh  --- change in PE

Vs = sqrt(10gh/7)

For disk

MI   Id = MR^2/2

KE = Id $\omega$²/2 +MV²/2 = (MR^2/2)*(V/R)^2 /2 + MV^2/2

= 3MV^2/4= Mgh

Vd = sqrt(4gh/3)

For the hoop

MI    Ih = MR^2

KE = Ih $\omega$²/2 +MV²/2 = (MR^2)*(V/R)^2 /2 + MV^2/2

      = 3MV^2/2 = Mgh

Vh = sqrt(2gh/3)